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David Howells108b42b2006-03-31 16:00:29 +01001 ============================
2 LINUX KERNEL MEMORY BARRIERS
3 ============================
4
5By: David Howells <dhowells@redhat.com>
David Howells90fddab2010-03-24 09:43:00 +00006 Paul E. McKenney <paulmck@linux.vnet.ibm.com>
Peter Zijlstrae7720af2016-04-26 10:22:05 -07007 Will Deacon <will.deacon@arm.com>
8 Peter Zijlstra <peterz@infradead.org>
David Howells108b42b2006-03-31 16:00:29 +01009
Peter Zijlstrae7720af2016-04-26 10:22:05 -070010==========
11DISCLAIMER
12==========
13
14This document is not a specification; it is intentionally (for the sake of
15brevity) and unintentionally (due to being human) incomplete. This document is
16meant as a guide to using the various memory barriers provided by Linux, but
Andrea Parri621df432018-02-20 15:25:07 -080017in case of any doubt (and there are many) please ask. Some doubts may be
18resolved by referring to the formal memory consistency model and related
19documentation at tools/memory-model/. Nevertheless, even this memory
20model should be viewed as the collective opinion of its maintainers rather
21than as an infallible oracle.
Peter Zijlstrae7720af2016-04-26 10:22:05 -070022
23To repeat, this document is not a specification of what Linux expects from
24hardware.
25
David Howells8d4840e2016-04-26 10:22:06 -070026The purpose of this document is twofold:
27
28 (1) to specify the minimum functionality that one can rely on for any
29 particular barrier, and
30
31 (2) to provide a guide as to how to use the barriers that are available.
32
33Note that an architecture can provide more than the minimum requirement
Stan Drozd35bdc722017-04-20 11:03:36 +020034for any particular barrier, but if the architecture provides less than
David Howells8d4840e2016-04-26 10:22:06 -070035that, that architecture is incorrect.
36
37Note also that it is possible that a barrier may be a no-op for an
38architecture because the way that arch works renders an explicit barrier
39unnecessary in that case.
40
41
Peter Zijlstrae7720af2016-04-26 10:22:05 -070042========
43CONTENTS
44========
David Howells108b42b2006-03-31 16:00:29 +010045
46 (*) Abstract memory access model.
47
48 - Device operations.
49 - Guarantees.
50
51 (*) What are memory barriers?
52
53 - Varieties of memory barrier.
54 - What may not be assumed about memory barriers?
Paul E. McKenneyf28f0862018-03-07 09:27:37 -080055 - Data dependency barriers (historical).
David Howells108b42b2006-03-31 16:00:29 +010056 - Control dependencies.
57 - SMP barrier pairing.
58 - Examples of memory barrier sequences.
David Howells670bd952006-06-10 09:54:12 -070059 - Read memory barriers vs load speculation.
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -070060 - Multicopy atomicity.
David Howells108b42b2006-03-31 16:00:29 +010061
62 (*) Explicit kernel barriers.
63
64 - Compiler barrier.
Jarek Poplawski81fc6322007-05-23 13:58:20 -070065 - CPU memory barriers.
David Howells108b42b2006-03-31 16:00:29 +010066 - MMIO write barrier.
67
68 (*) Implicit kernel memory barriers.
69
SeongJae Park166bda72016-04-12 08:52:50 -070070 - Lock acquisition functions.
David Howells108b42b2006-03-31 16:00:29 +010071 - Interrupt disabling functions.
David Howells50fa6102009-04-28 15:01:38 +010072 - Sleep and wake-up functions.
David Howells108b42b2006-03-31 16:00:29 +010073 - Miscellaneous functions.
74
SeongJae Park166bda72016-04-12 08:52:50 -070075 (*) Inter-CPU acquiring barrier effects.
David Howells108b42b2006-03-31 16:00:29 +010076
SeongJae Park166bda72016-04-12 08:52:50 -070077 - Acquires vs memory accesses.
78 - Acquires vs I/O accesses.
David Howells108b42b2006-03-31 16:00:29 +010079
80 (*) Where are memory barriers needed?
81
82 - Interprocessor interaction.
83 - Atomic operations.
84 - Accessing devices.
85 - Interrupts.
86
87 (*) Kernel I/O barrier effects.
88
89 (*) Assumed minimum execution ordering model.
90
91 (*) The effects of the cpu cache.
92
93 - Cache coherency.
94 - Cache coherency vs DMA.
95 - Cache coherency vs MMIO.
96
97 (*) The things CPUs get up to.
98
99 - And then there's the Alpha.
SeongJae Park01e1cd62016-04-12 08:52:51 -0700100 - Virtual Machine Guests.
David Howells108b42b2006-03-31 16:00:29 +0100101
David Howells90fddab2010-03-24 09:43:00 +0000102 (*) Example uses.
103
104 - Circular buffers.
105
David Howells108b42b2006-03-31 16:00:29 +0100106 (*) References.
107
108
109============================
110ABSTRACT MEMORY ACCESS MODEL
111============================
112
113Consider the following abstract model of the system:
114
115 : :
116 : :
117 : :
118 +-------+ : +--------+ : +-------+
119 | | : | | : | |
120 | | : | | : | |
121 | CPU 1 |<----->| Memory |<----->| CPU 2 |
122 | | : | | : | |
123 | | : | | : | |
124 +-------+ : +--------+ : +-------+
125 ^ : ^ : ^
126 | : | : |
127 | : | : |
128 | : v : |
129 | : +--------+ : |
130 | : | | : |
131 | : | | : |
132 +---------->| Device |<----------+
133 : | | :
134 : | | :
135 : +--------+ :
136 : :
137
138Each CPU executes a program that generates memory access operations. In the
139abstract CPU, memory operation ordering is very relaxed, and a CPU may actually
140perform the memory operations in any order it likes, provided program causality
141appears to be maintained. Similarly, the compiler may also arrange the
142instructions it emits in any order it likes, provided it doesn't affect the
143apparent operation of the program.
144
145So in the above diagram, the effects of the memory operations performed by a
146CPU are perceived by the rest of the system as the operations cross the
147interface between the CPU and rest of the system (the dotted lines).
148
149
150For example, consider the following sequence of events:
151
152 CPU 1 CPU 2
153 =============== ===============
154 { A == 1; B == 2 }
Alexey Dobriyan615cc2c2014-06-06 14:36:41 -0700155 A = 3; x = B;
156 B = 4; y = A;
David Howells108b42b2006-03-31 16:00:29 +0100157
158The set of accesses as seen by the memory system in the middle can be arranged
159in 24 different combinations:
160
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400161 STORE A=3, STORE B=4, y=LOAD A->3, x=LOAD B->4
162 STORE A=3, STORE B=4, x=LOAD B->4, y=LOAD A->3
163 STORE A=3, y=LOAD A->3, STORE B=4, x=LOAD B->4
164 STORE A=3, y=LOAD A->3, x=LOAD B->2, STORE B=4
165 STORE A=3, x=LOAD B->2, STORE B=4, y=LOAD A->3
166 STORE A=3, x=LOAD B->2, y=LOAD A->3, STORE B=4
167 STORE B=4, STORE A=3, y=LOAD A->3, x=LOAD B->4
David Howells108b42b2006-03-31 16:00:29 +0100168 STORE B=4, ...
169 ...
170
171and can thus result in four different combinations of values:
172
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400173 x == 2, y == 1
174 x == 2, y == 3
175 x == 4, y == 1
176 x == 4, y == 3
David Howells108b42b2006-03-31 16:00:29 +0100177
178
179Furthermore, the stores committed by a CPU to the memory system may not be
180perceived by the loads made by another CPU in the same order as the stores were
181committed.
182
183
184As a further example, consider this sequence of events:
185
186 CPU 1 CPU 2
187 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700188 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100189 B = 4; Q = P;
190 P = &B D = *Q;
191
192There is an obvious data dependency here, as the value loaded into D depends on
193the address retrieved from P by CPU 2. At the end of the sequence, any of the
194following results are possible:
195
196 (Q == &A) and (D == 1)
197 (Q == &B) and (D == 2)
198 (Q == &B) and (D == 4)
199
200Note that CPU 2 will never try and load C into D because the CPU will load P
201into Q before issuing the load of *Q.
202
203
204DEVICE OPERATIONS
205-----------------
206
207Some devices present their control interfaces as collections of memory
208locations, but the order in which the control registers are accessed is very
209important. For instance, imagine an ethernet card with a set of internal
210registers that are accessed through an address port register (A) and a data
211port register (D). To read internal register 5, the following code might then
212be used:
213
214 *A = 5;
215 x = *D;
216
217but this might show up as either of the following two sequences:
218
219 STORE *A = 5, x = LOAD *D
220 x = LOAD *D, STORE *A = 5
221
222the second of which will almost certainly result in a malfunction, since it set
223the address _after_ attempting to read the register.
224
225
226GUARANTEES
227----------
228
229There are some minimal guarantees that may be expected of a CPU:
230
231 (*) On any given CPU, dependent memory accesses will be issued in order, with
232 respect to itself. This means that for:
233
Paul E. McKenney40555942017-10-09 09:15:21 -0700234 Q = READ_ONCE(P); D = READ_ONCE(*Q);
David Howells108b42b2006-03-31 16:00:29 +0100235
236 the CPU will issue the following memory operations:
237
238 Q = LOAD P, D = LOAD *Q
239
Paul E. McKenney40555942017-10-09 09:15:21 -0700240 and always in that order. However, on DEC Alpha, READ_ONCE() also
241 emits a memory-barrier instruction, so that a DEC Alpha CPU will
242 instead issue the following memory operations:
243
244 Q = LOAD P, MEMORY_BARRIER, D = LOAD *Q, MEMORY_BARRIER
245
246 Whether on DEC Alpha or not, the READ_ONCE() also prevents compiler
247 mischief.
David Howells108b42b2006-03-31 16:00:29 +0100248
249 (*) Overlapping loads and stores within a particular CPU will appear to be
250 ordered within that CPU. This means that for:
251
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700252 a = READ_ONCE(*X); WRITE_ONCE(*X, b);
David Howells108b42b2006-03-31 16:00:29 +0100253
254 the CPU will only issue the following sequence of memory operations:
255
256 a = LOAD *X, STORE *X = b
257
258 And for:
259
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700260 WRITE_ONCE(*X, c); d = READ_ONCE(*X);
David Howells108b42b2006-03-31 16:00:29 +0100261
262 the CPU will only issue:
263
264 STORE *X = c, d = LOAD *X
265
Matt LaPlantefa00e7e2006-11-30 04:55:36 +0100266 (Loads and stores overlap if they are targeted at overlapping pieces of
David Howells108b42b2006-03-31 16:00:29 +0100267 memory).
268
269And there are a number of things that _must_ or _must_not_ be assumed:
270
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700271 (*) It _must_not_ be assumed that the compiler will do what you want
272 with memory references that are not protected by READ_ONCE() and
273 WRITE_ONCE(). Without them, the compiler is within its rights to
274 do all sorts of "creative" transformations, which are covered in
Paul E. McKenney895f5542016-01-06 14:23:03 -0800275 the COMPILER BARRIER section.
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800276
David Howells108b42b2006-03-31 16:00:29 +0100277 (*) It _must_not_ be assumed that independent loads and stores will be issued
278 in the order given. This means that for:
279
280 X = *A; Y = *B; *D = Z;
281
282 we may get any of the following sequences:
283
284 X = LOAD *A, Y = LOAD *B, STORE *D = Z
285 X = LOAD *A, STORE *D = Z, Y = LOAD *B
286 Y = LOAD *B, X = LOAD *A, STORE *D = Z
287 Y = LOAD *B, STORE *D = Z, X = LOAD *A
288 STORE *D = Z, X = LOAD *A, Y = LOAD *B
289 STORE *D = Z, Y = LOAD *B, X = LOAD *A
290
291 (*) It _must_ be assumed that overlapping memory accesses may be merged or
292 discarded. This means that for:
293
294 X = *A; Y = *(A + 4);
295
296 we may get any one of the following sequences:
297
298 X = LOAD *A; Y = LOAD *(A + 4);
299 Y = LOAD *(A + 4); X = LOAD *A;
300 {X, Y} = LOAD {*A, *(A + 4) };
301
302 And for:
303
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700304 *A = X; *(A + 4) = Y;
David Howells108b42b2006-03-31 16:00:29 +0100305
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700306 we may get any of:
David Howells108b42b2006-03-31 16:00:29 +0100307
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700308 STORE *A = X; STORE *(A + 4) = Y;
309 STORE *(A + 4) = Y; STORE *A = X;
310 STORE {*A, *(A + 4) } = {X, Y};
David Howells108b42b2006-03-31 16:00:29 +0100311
Paul E. McKenney432fbf32014-09-04 17:12:49 -0700312And there are anti-guarantees:
313
314 (*) These guarantees do not apply to bitfields, because compilers often
315 generate code to modify these using non-atomic read-modify-write
316 sequences. Do not attempt to use bitfields to synchronize parallel
317 algorithms.
318
319 (*) Even in cases where bitfields are protected by locks, all fields
320 in a given bitfield must be protected by one lock. If two fields
321 in a given bitfield are protected by different locks, the compiler's
322 non-atomic read-modify-write sequences can cause an update to one
323 field to corrupt the value of an adjacent field.
324
325 (*) These guarantees apply only to properly aligned and sized scalar
326 variables. "Properly sized" currently means variables that are
327 the same size as "char", "short", "int" and "long". "Properly
328 aligned" means the natural alignment, thus no constraints for
329 "char", two-byte alignment for "short", four-byte alignment for
330 "int", and either four-byte or eight-byte alignment for "long",
331 on 32-bit and 64-bit systems, respectively. Note that these
332 guarantees were introduced into the C11 standard, so beware when
333 using older pre-C11 compilers (for example, gcc 4.6). The portion
334 of the standard containing this guarantee is Section 3.14, which
335 defines "memory location" as follows:
336
337 memory location
338 either an object of scalar type, or a maximal sequence
339 of adjacent bit-fields all having nonzero width
340
341 NOTE 1: Two threads of execution can update and access
342 separate memory locations without interfering with
343 each other.
344
345 NOTE 2: A bit-field and an adjacent non-bit-field member
346 are in separate memory locations. The same applies
347 to two bit-fields, if one is declared inside a nested
348 structure declaration and the other is not, or if the two
349 are separated by a zero-length bit-field declaration,
350 or if they are separated by a non-bit-field member
351 declaration. It is not safe to concurrently update two
352 bit-fields in the same structure if all members declared
353 between them are also bit-fields, no matter what the
354 sizes of those intervening bit-fields happen to be.
355
David Howells108b42b2006-03-31 16:00:29 +0100356
357=========================
358WHAT ARE MEMORY BARRIERS?
359=========================
360
361As can be seen above, independent memory operations are effectively performed
362in random order, but this can be a problem for CPU-CPU interaction and for I/O.
363What is required is some way of intervening to instruct the compiler and the
364CPU to restrict the order.
365
366Memory barriers are such interventions. They impose a perceived partial
David Howells2b948952006-06-25 05:48:49 -0700367ordering over the memory operations on either side of the barrier.
368
369Such enforcement is important because the CPUs and other devices in a system
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700370can use a variety of tricks to improve performance, including reordering,
David Howells2b948952006-06-25 05:48:49 -0700371deferral and combination of memory operations; speculative loads; speculative
372branch prediction and various types of caching. Memory barriers are used to
373override or suppress these tricks, allowing the code to sanely control the
374interaction of multiple CPUs and/or devices.
David Howells108b42b2006-03-31 16:00:29 +0100375
376
377VARIETIES OF MEMORY BARRIER
378---------------------------
379
380Memory barriers come in four basic varieties:
381
382 (1) Write (or store) memory barriers.
383
384 A write memory barrier gives a guarantee that all the STORE operations
385 specified before the barrier will appear to happen before all the STORE
386 operations specified after the barrier with respect to the other
387 components of the system.
388
389 A write barrier is a partial ordering on stores only; it is not required
390 to have any effect on loads.
391
David Howells6bc39272006-06-25 05:49:22 -0700392 A CPU can be viewed as committing a sequence of store operations to the
Guilherme G. Piccoli5692fcc2017-09-21 16:29:01 -0300393 memory system as time progresses. All stores _before_ a write barrier
394 will occur _before_ all the stores after the write barrier.
David Howells108b42b2006-03-31 16:00:29 +0100395
396 [!] Note that write barriers should normally be paired with read or data
397 dependency barriers; see the "SMP barrier pairing" subsection.
398
399
400 (2) Data dependency barriers.
401
402 A data dependency barrier is a weaker form of read barrier. In the case
403 where two loads are performed such that the second depends on the result
404 of the first (eg: the first load retrieves the address to which the second
405 load will be directed), a data dependency barrier would be required to
Nikolay Borisov51de7882018-02-20 15:25:08 -0800406 make sure that the target of the second load is updated after the address
David Howells108b42b2006-03-31 16:00:29 +0100407 obtained by the first load is accessed.
408
409 A data dependency barrier is a partial ordering on interdependent loads
410 only; it is not required to have any effect on stores, independent loads
411 or overlapping loads.
412
413 As mentioned in (1), the other CPUs in the system can be viewed as
414 committing sequences of stores to the memory system that the CPU being
415 considered can then perceive. A data dependency barrier issued by the CPU
416 under consideration guarantees that for any load preceding it, if that
417 load touches one of a sequence of stores from another CPU, then by the
418 time the barrier completes, the effects of all the stores prior to that
419 touched by the load will be perceptible to any loads issued after the data
420 dependency barrier.
421
422 See the "Examples of memory barrier sequences" subsection for diagrams
423 showing the ordering constraints.
424
425 [!] Note that the first load really has to have a _data_ dependency and
426 not a control dependency. If the address for the second load is dependent
427 on the first load, but the dependency is through a conditional rather than
428 actually loading the address itself, then it's a _control_ dependency and
429 a full read barrier or better is required. See the "Control dependencies"
430 subsection for more information.
431
432 [!] Note that data dependency barriers should normally be paired with
433 write barriers; see the "SMP barrier pairing" subsection.
434
435
436 (3) Read (or load) memory barriers.
437
438 A read barrier is a data dependency barrier plus a guarantee that all the
439 LOAD operations specified before the barrier will appear to happen before
440 all the LOAD operations specified after the barrier with respect to the
441 other components of the system.
442
443 A read barrier is a partial ordering on loads only; it is not required to
444 have any effect on stores.
445
446 Read memory barriers imply data dependency barriers, and so can substitute
447 for them.
448
449 [!] Note that read barriers should normally be paired with write barriers;
450 see the "SMP barrier pairing" subsection.
451
452
453 (4) General memory barriers.
454
David Howells670bd952006-06-10 09:54:12 -0700455 A general memory barrier gives a guarantee that all the LOAD and STORE
456 operations specified before the barrier will appear to happen before all
457 the LOAD and STORE operations specified after the barrier with respect to
458 the other components of the system.
459
460 A general memory barrier is a partial ordering over both loads and stores.
David Howells108b42b2006-03-31 16:00:29 +0100461
462 General memory barriers imply both read and write memory barriers, and so
463 can substitute for either.
464
465
466And a couple of implicit varieties:
467
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100468 (5) ACQUIRE operations.
David Howells108b42b2006-03-31 16:00:29 +0100469
470 This acts as a one-way permeable barrier. It guarantees that all memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100471 operations after the ACQUIRE operation will appear to happen after the
472 ACQUIRE operation with respect to the other components of the system.
Davidlohr Bueso787df632016-04-12 08:52:55 -0700473 ACQUIRE operations include LOCK operations and both smp_load_acquire()
474 and smp_cond_acquire() operations. The later builds the necessary ACQUIRE
475 semantics from relying on a control dependency and smp_rmb().
David Howells108b42b2006-03-31 16:00:29 +0100476
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100477 Memory operations that occur before an ACQUIRE operation may appear to
478 happen after it completes.
David Howells108b42b2006-03-31 16:00:29 +0100479
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100480 An ACQUIRE operation should almost always be paired with a RELEASE
481 operation.
David Howells108b42b2006-03-31 16:00:29 +0100482
483
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100484 (6) RELEASE operations.
David Howells108b42b2006-03-31 16:00:29 +0100485
486 This also acts as a one-way permeable barrier. It guarantees that all
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100487 memory operations before the RELEASE operation will appear to happen
488 before the RELEASE operation with respect to the other components of the
489 system. RELEASE operations include UNLOCK operations and
490 smp_store_release() operations.
David Howells108b42b2006-03-31 16:00:29 +0100491
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100492 Memory operations that occur after a RELEASE operation may appear to
David Howells108b42b2006-03-31 16:00:29 +0100493 happen before it completes.
494
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100495 The use of ACQUIRE and RELEASE operations generally precludes the need
496 for other sorts of memory barrier (but note the exceptions mentioned in
497 the subsection "MMIO write barrier"). In addition, a RELEASE+ACQUIRE
498 pair is -not- guaranteed to act as a full memory barrier. However, after
499 an ACQUIRE on a given variable, all memory accesses preceding any prior
500 RELEASE on that same variable are guaranteed to be visible. In other
501 words, within a given variable's critical section, all accesses of all
502 previous critical sections for that variable are guaranteed to have
503 completed.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -0800504
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100505 This means that ACQUIRE acts as a minimal "acquire" operation and
506 RELEASE acts as a minimal "release" operation.
David Howells108b42b2006-03-31 16:00:29 +0100507
Peter Zijlstra706eeb32017-06-12 14:50:27 +0200508A subset of the atomic operations described in atomic_t.txt have ACQUIRE and
509RELEASE variants in addition to fully-ordered and relaxed (no barrier
510semantics) definitions. For compound atomics performing both a load and a
511store, ACQUIRE semantics apply only to the load and RELEASE semantics apply
512only to the store portion of the operation.
David Howells108b42b2006-03-31 16:00:29 +0100513
514Memory barriers are only required where there's a possibility of interaction
515between two CPUs or between a CPU and a device. If it can be guaranteed that
516there won't be any such interaction in any particular piece of code, then
517memory barriers are unnecessary in that piece of code.
518
519
520Note that these are the _minimum_ guarantees. Different architectures may give
521more substantial guarantees, but they may _not_ be relied upon outside of arch
522specific code.
523
524
525WHAT MAY NOT BE ASSUMED ABOUT MEMORY BARRIERS?
526----------------------------------------------
527
528There are certain things that the Linux kernel memory barriers do not guarantee:
529
530 (*) There is no guarantee that any of the memory accesses specified before a
531 memory barrier will be _complete_ by the completion of a memory barrier
532 instruction; the barrier can be considered to draw a line in that CPU's
533 access queue that accesses of the appropriate type may not cross.
534
535 (*) There is no guarantee that issuing a memory barrier on one CPU will have
536 any direct effect on another CPU or any other hardware in the system. The
537 indirect effect will be the order in which the second CPU sees the effects
538 of the first CPU's accesses occur, but see the next point:
539
David Howells6bc39272006-06-25 05:49:22 -0700540 (*) There is no guarantee that a CPU will see the correct order of effects
David Howells108b42b2006-03-31 16:00:29 +0100541 from a second CPU's accesses, even _if_ the second CPU uses a memory
542 barrier, unless the first CPU _also_ uses a matching memory barrier (see
543 the subsection on "SMP Barrier Pairing").
544
545 (*) There is no guarantee that some intervening piece of off-the-CPU
546 hardware[*] will not reorder the memory accesses. CPU cache coherency
547 mechanisms should propagate the indirect effects of a memory barrier
548 between CPUs, but might not do so in order.
549
550 [*] For information on bus mastering DMA and coherency please read:
551
Randy Dunlap4b5ff462008-03-10 17:16:32 -0700552 Documentation/PCI/pci.txt
Paul Bolle395cf962011-08-15 02:02:26 +0200553 Documentation/DMA-API-HOWTO.txt
David Howells108b42b2006-03-31 16:00:29 +0100554 Documentation/DMA-API.txt
555
556
Paul E. McKenneyf28f0862018-03-07 09:27:37 -0800557DATA DEPENDENCY BARRIERS (HISTORICAL)
558-------------------------------------
559
560As of v4.15 of the Linux kernel, an smp_read_barrier_depends() was
561added to READ_ONCE(), which means that about the only people who
562need to pay attention to this section are those working on DEC Alpha
563architecture-specific code and those working on READ_ONCE() itself.
564For those who need it, and for those who are interested in the history,
565here is the story of data-dependency barriers.
David Howells108b42b2006-03-31 16:00:29 +0100566
567The usage requirements of data dependency barriers are a little subtle, and
568it's not always obvious that they're needed. To illustrate, consider the
569following sequence of events:
570
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800571 CPU 1 CPU 2
572 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700573 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100574 B = 4;
575 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700576 WRITE_ONCE(P, &B)
577 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800578 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100579
580There's a clear data dependency here, and it would seem that by the end of the
581sequence, Q must be either &A or &B, and that:
582
583 (Q == &A) implies (D == 1)
584 (Q == &B) implies (D == 4)
585
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700586But! CPU 2's perception of P may be updated _before_ its perception of B, thus
David Howells108b42b2006-03-31 16:00:29 +0100587leading to the following situation:
588
589 (Q == &B) and (D == 2) ????
590
591Whilst this may seem like a failure of coherency or causality maintenance, it
592isn't, and this behaviour can be observed on certain real CPUs (such as the DEC
593Alpha).
594
David Howells2b948952006-06-25 05:48:49 -0700595To deal with this, a data dependency barrier or better must be inserted
596between the address load and the data load:
David Howells108b42b2006-03-31 16:00:29 +0100597
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800598 CPU 1 CPU 2
599 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700600 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100601 B = 4;
602 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700603 WRITE_ONCE(P, &B);
604 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800605 <data dependency barrier>
606 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100607
608This enforces the occurrence of one of the two implications, and prevents the
609third possibility from arising.
610
Paul E. McKenney92a84dd2016-01-14 14:17:04 -0800611
David Howells108b42b2006-03-31 16:00:29 +0100612[!] Note that this extremely counterintuitive situation arises most easily on
613machines with split caches, so that, for example, one cache bank processes
614even-numbered cache lines and the other bank processes odd-numbered cache
615lines. The pointer P might be stored in an odd-numbered cache line, and the
616variable B might be stored in an even-numbered cache line. Then, if the
617even-numbered bank of the reading CPU's cache is extremely busy while the
618odd-numbered bank is idle, one can see the new value of the pointer P (&B),
David Howells6bc39272006-06-25 05:49:22 -0700619but the old value of the variable B (2).
David Howells108b42b2006-03-31 16:00:29 +0100620
621
Paul E. McKenney66ce3a42017-06-30 16:18:28 -0700622A data-dependency barrier is not required to order dependent writes
623because the CPUs that the Linux kernel supports don't do writes
624until they are certain (1) that the write will actually happen, (2)
625of the location of the write, and (3) of the value to be written.
626But please carefully read the "CONTROL DEPENDENCIES" section and the
627Documentation/RCU/rcu_dereference.txt file: The compiler can and does
628break dependencies in a great many highly creative ways.
629
630 CPU 1 CPU 2
631 =============== ===============
632 { A == 1, B == 2, C = 3, P == &A, Q == &C }
633 B = 4;
634 <write barrier>
635 WRITE_ONCE(P, &B);
636 Q = READ_ONCE(P);
637 WRITE_ONCE(*Q, 5);
638
639Therefore, no data-dependency barrier is required to order the read into
640Q with the store into *Q. In other words, this outcome is prohibited,
641even without a data-dependency barrier:
642
643 (Q == &B) && (B == 4)
644
645Please note that this pattern should be rare. After all, the whole point
646of dependency ordering is to -prevent- writes to the data structure, along
647with the expensive cache misses associated with those writes. This pattern
648can be used to record rare error conditions and the like, and the CPUs'
649naturally occurring ordering prevents such records from being lost.
650
651
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700652Note well that the ordering provided by a data dependency is local to
653the CPU containing it. See the section on "Multicopy atomicity" for
654more information.
655
656
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800657The data dependency barrier is very important to the RCU system,
658for example. See rcu_assign_pointer() and rcu_dereference() in
659include/linux/rcupdate.h. This permits the current target of an RCU'd
660pointer to be replaced with a new modified target, without the replacement
661target appearing to be incompletely initialised.
David Howells108b42b2006-03-31 16:00:29 +0100662
663See also the subsection on "Cache Coherency" for a more thorough example.
664
665
666CONTROL DEPENDENCIES
667--------------------
668
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800669Control dependencies can be a bit tricky because current compilers do
670not understand them. The purpose of this section is to help you prevent
671the compiler's ignorance from breaking your code.
672
Paul E. McKenneyff382812015-02-17 10:00:06 -0800673A load-load control dependency requires a full read memory barrier, not
674simply a data dependency barrier to make it work correctly. Consider the
675following bit of code:
David Howells108b42b2006-03-31 16:00:29 +0100676
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700677 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800678 if (q) {
679 <data dependency barrier> /* BUG: No data dependency!!! */
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700680 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700681 }
David Howells108b42b2006-03-31 16:00:29 +0100682
683This will not have the desired effect because there is no actual data
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800684dependency, but rather a control dependency that the CPU may short-circuit
685by attempting to predict the outcome in advance, so that other CPUs see
686the load from b as having happened before the load from a. In such a
687case what's actually required is:
David Howells108b42b2006-03-31 16:00:29 +0100688
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700689 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800690 if (q) {
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700691 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700692 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700693 }
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800694
695However, stores are not speculated. This means that ordering -is- provided
Paul E. McKenneyff382812015-02-17 10:00:06 -0800696for load-store control dependencies, as in the following example:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800697
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800698 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700699 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800700 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800701 }
702
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800703Control dependencies pair normally with other types of barriers.
704That said, please note that neither READ_ONCE() nor WRITE_ONCE()
705are optional! Without the READ_ONCE(), the compiler might combine the
706load from 'a' with other loads from 'a'. Without the WRITE_ONCE(),
707the compiler might combine the store to 'b' with other stores to 'b'.
708Either can result in highly counterintuitive effects on ordering.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800709
710Worse yet, if the compiler is able to prove (say) that the value of
711variable 'a' is always non-zero, it would be well within its rights
712to optimize the original example by eliminating the "if" statement
713as follows:
714
715 q = a;
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800716 b = 1; /* BUG: Compiler and CPU can both reorder!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800717
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800718So don't leave out the READ_ONCE().
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700719
720It is tempting to try to enforce ordering on identical stores on both
721branches of the "if" statement as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800722
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800723 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800724 if (q) {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800725 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800726 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800727 do_something();
728 } else {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800729 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800730 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800731 do_something_else();
732 }
733
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700734Unfortunately, current compilers will transform this as follows at high
735optimization levels:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800736
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800737 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700738 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800739 WRITE_ONCE(b, 1); /* BUG: No ordering vs. load from a!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800740 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800741 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800742 do_something();
743 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800744 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800745 do_something_else();
746 }
747
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700748Now there is no conditional between the load from 'a' and the store to
749'b', which means that the CPU is within its rights to reorder them:
750The conditional is absolutely required, and must be present in the
751assembly code even after all compiler optimizations have been applied.
752Therefore, if you need ordering in this example, you need explicit
753memory barriers, for example, smp_store_release():
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800754
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700755 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700756 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800757 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800758 do_something();
759 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800760 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800761 do_something_else();
762 }
763
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700764In contrast, without explicit memory barriers, two-legged-if control
765ordering is guaranteed only when the stores differ, for example:
766
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800767 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700768 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800769 WRITE_ONCE(b, 1);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700770 do_something();
771 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800772 WRITE_ONCE(b, 2);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700773 do_something_else();
774 }
775
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800776The initial READ_ONCE() is still required to prevent the compiler from
777proving the value of 'a'.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800778
779In addition, you need to be careful what you do with the local variable 'q',
780otherwise the compiler might be able to guess the value and again remove
781the needed conditional. For example:
782
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800783 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800784 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800785 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800786 do_something();
787 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800788 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800789 do_something_else();
790 }
791
792If MAX is defined to be 1, then the compiler knows that (q % MAX) is
793equal to zero, in which case the compiler is within its rights to
794transform the above code into the following:
795
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800796 q = READ_ONCE(a);
pierre Kuob26cfc42017-04-07 14:37:36 +0800797 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800798 do_something_else();
799
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700800Given this transformation, the CPU is not required to respect the ordering
801between the load from variable 'a' and the store to variable 'b'. It is
802tempting to add a barrier(), but this does not help. The conditional
803is gone, and the barrier won't bring it back. Therefore, if you are
804relying on this ordering, you should make sure that MAX is greater than
805one, perhaps as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800806
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800807 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800808 BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
809 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800810 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800811 do_something();
812 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800813 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800814 do_something_else();
815 }
816
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700817Please note once again that the stores to 'b' differ. If they were
818identical, as noted earlier, the compiler could pull this store outside
819of the 'if' statement.
820
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700821You must also be careful not to rely too much on boolean short-circuit
822evaluation. Consider this example:
823
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800824 q = READ_ONCE(a);
Paul E. McKenney57aecae2015-05-18 18:27:42 -0700825 if (q || 1 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700826 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700827
Paul E. McKenney5af46922015-04-25 12:48:29 -0700828Because the first condition cannot fault and the second condition is
829always true, the compiler can transform this example as following,
830defeating control dependency:
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700831
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800832 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700833 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700834
835This example underscores the need to ensure that the compiler cannot
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700836out-guess your code. More generally, although READ_ONCE() does force
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700837the compiler to actually emit code for a given load, it does not force
838the compiler to use the results.
839
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700840In addition, control dependencies apply only to the then-clause and
841else-clause of the if-statement in question. In particular, it does
842not necessarily apply to code following the if-statement:
843
844 q = READ_ONCE(a);
845 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800846 WRITE_ONCE(b, 1);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700847 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800848 WRITE_ONCE(b, 2);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700849 }
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800850 WRITE_ONCE(c, 1); /* BUG: No ordering against the read from 'a'. */
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700851
852It is tempting to argue that there in fact is ordering because the
853compiler cannot reorder volatile accesses and also cannot reorder
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800854the writes to 'b' with the condition. Unfortunately for this line
855of reasoning, the compiler might compile the two writes to 'b' as
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700856conditional-move instructions, as in this fanciful pseudo-assembly
857language:
858
859 ld r1,a
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700860 cmp r1,$0
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800861 cmov,ne r4,$1
862 cmov,eq r4,$2
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700863 st r4,b
864 st $1,c
865
866A weakly ordered CPU would have no dependency of any sort between the load
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800867from 'a' and the store to 'c'. The control dependencies would extend
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700868only to the pair of cmov instructions and the store depending on them.
869In short, control dependencies apply only to the stores in the then-clause
870and else-clause of the if-statement in question (including functions
871invoked by those two clauses), not to code following that if-statement.
872
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800873
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700874Note well that the ordering provided by a control dependency is local
875to the CPU containing it. See the section on "Multicopy atomicity"
876for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800877
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800878
879In summary:
880
881 (*) Control dependencies can order prior loads against later stores.
882 However, they do -not- guarantee any other sort of ordering:
883 Not prior loads against later loads, nor prior stores against
884 later anything. If you need these other forms of ordering,
Davidlohr Buesod87510c2014-12-28 01:11:16 -0800885 use smp_rmb(), smp_wmb(), or, in the case of prior stores and
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800886 later loads, smp_mb().
887
Paul E. McKenney7817b792015-12-29 16:23:18 -0800888 (*) If both legs of the "if" statement begin with identical stores to
889 the same variable, then those stores must be ordered, either by
890 preceding both of them with smp_mb() or by using smp_store_release()
891 to carry out the stores. Please note that it is -not- sufficient
Paul E. McKenneya5052652016-04-12 08:52:49 -0700892 to use barrier() at beginning of each leg of the "if" statement
893 because, as shown by the example above, optimizing compilers can
894 destroy the control dependency while respecting the letter of the
895 barrier() law.
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800896
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800897 (*) Control dependencies require at least one run-time conditional
Paul E. McKenney586dd562014-02-11 12:28:06 -0800898 between the prior load and the subsequent store, and this
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700899 conditional must involve the prior load. If the compiler is able
900 to optimize the conditional away, it will have also optimized
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800901 away the ordering. Careful use of READ_ONCE() and WRITE_ONCE()
902 can help to preserve the needed conditional.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800903
904 (*) Control dependencies require that the compiler avoid reordering the
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800905 dependency into nonexistence. Careful use of READ_ONCE() or
906 atomic{,64}_read() can help to preserve your control dependency.
Paul E. McKenney895f5542016-01-06 14:23:03 -0800907 Please see the COMPILER BARRIER section for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800908
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700909 (*) Control dependencies apply only to the then-clause and else-clause
910 of the if-statement containing the control dependency, including
911 any functions that these two clauses call. Control dependencies
912 do -not- apply to code following the if-statement containing the
913 control dependency.
914
Paul E. McKenneyff382812015-02-17 10:00:06 -0800915 (*) Control dependencies pair normally with other types of barriers.
916
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700917 (*) Control dependencies do -not- provide multicopy atomicity. If you
918 need all the CPUs to see a given store at the same time, use smp_mb().
David Howells108b42b2006-03-31 16:00:29 +0100919
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800920 (*) Compilers do not understand control dependencies. It is therefore
921 your job to ensure that they do not break your code.
922
David Howells108b42b2006-03-31 16:00:29 +0100923
924SMP BARRIER PAIRING
925-------------------
926
927When dealing with CPU-CPU interactions, certain types of memory barrier should
928always be paired. A lack of appropriate pairing is almost certainly an error.
929
Paul E. McKenneyff382812015-02-17 10:00:06 -0800930General barriers pair with each other, though they also pair with most
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700931other types of barriers, albeit without multicopy atomicity. An acquire
932barrier pairs with a release barrier, but both may also pair with other
933barriers, including of course general barriers. A write barrier pairs
934with a data dependency barrier, a control dependency, an acquire barrier,
935a release barrier, a read barrier, or a general barrier. Similarly a
936read barrier, control dependency, or a data dependency barrier pairs
937with a write barrier, an acquire barrier, a release barrier, or a
938general barrier:
David Howells108b42b2006-03-31 16:00:29 +0100939
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800940 CPU 1 CPU 2
941 =============== ===============
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700942 WRITE_ONCE(a, 1);
David Howells108b42b2006-03-31 16:00:29 +0100943 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700944 WRITE_ONCE(b, 2); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800945 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700946 y = READ_ONCE(a);
David Howells108b42b2006-03-31 16:00:29 +0100947
948Or:
949
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800950 CPU 1 CPU 2
951 =============== ===============================
David Howells108b42b2006-03-31 16:00:29 +0100952 a = 1;
953 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700954 WRITE_ONCE(b, &a); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800955 <data dependency barrier>
956 y = *x;
David Howells108b42b2006-03-31 16:00:29 +0100957
Paul E. McKenneyff382812015-02-17 10:00:06 -0800958Or even:
959
960 CPU 1 CPU 2
961 =============== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700962 r1 = READ_ONCE(y);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800963 <general barrier>
Scott Tsaid92f8422017-09-20 02:16:00 +0800964 WRITE_ONCE(x, 1); if (r2 = READ_ONCE(x)) {
Paul E. McKenneyff382812015-02-17 10:00:06 -0800965 <implicit control dependency>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700966 WRITE_ONCE(y, 1);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800967 }
968
969 assert(r1 == 0 || r2 == 0);
970
David Howells108b42b2006-03-31 16:00:29 +0100971Basically, the read barrier always has to be there, even though it can be of
972the "weaker" type.
973
David Howells670bd952006-06-10 09:54:12 -0700974[!] Note that the stores before the write barrier would normally be expected to
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700975match the loads after the read barrier or the data dependency barrier, and vice
David Howells670bd952006-06-10 09:54:12 -0700976versa:
977
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800978 CPU 1 CPU 2
979 =================== ===================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700980 WRITE_ONCE(a, 1); }---- --->{ v = READ_ONCE(c);
981 WRITE_ONCE(b, 2); } \ / { w = READ_ONCE(d);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800982 <write barrier> \ <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700983 WRITE_ONCE(c, 3); } / \ { x = READ_ONCE(a);
984 WRITE_ONCE(d, 4); }---- --->{ y = READ_ONCE(b);
David Howells670bd952006-06-10 09:54:12 -0700985
David Howells108b42b2006-03-31 16:00:29 +0100986
987EXAMPLES OF MEMORY BARRIER SEQUENCES
988------------------------------------
989
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700990Firstly, write barriers act as partial orderings on store operations.
David Howells108b42b2006-03-31 16:00:29 +0100991Consider the following sequence of events:
992
993 CPU 1
994 =======================
995 STORE A = 1
996 STORE B = 2
997 STORE C = 3
998 <write barrier>
999 STORE D = 4
1000 STORE E = 5
1001
1002This sequence of events is committed to the memory coherence system in an order
1003that the rest of the system might perceive as the unordered set of { STORE A,
Adrian Bunk80f72282006-06-30 18:27:16 +02001004STORE B, STORE C } all occurring before the unordered set of { STORE D, STORE E
David Howells108b42b2006-03-31 16:00:29 +01001005}:
1006
1007 +-------+ : :
1008 | | +------+
1009 | |------>| C=3 | } /\
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001010 | | : +------+ }----- \ -----> Events perceptible to
1011 | | : | A=1 | } \/ the rest of the system
David Howells108b42b2006-03-31 16:00:29 +01001012 | | : +------+ }
1013 | CPU 1 | : | B=2 | }
1014 | | +------+ }
1015 | | wwwwwwwwwwwwwwww } <--- At this point the write barrier
1016 | | +------+ } requires all stores prior to the
1017 | | : | E=5 | } barrier to be committed before
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001018 | | : +------+ } further stores may take place
David Howells108b42b2006-03-31 16:00:29 +01001019 | |------>| D=4 | }
1020 | | +------+
1021 +-------+ : :
1022 |
David Howells670bd952006-06-10 09:54:12 -07001023 | Sequence in which stores are committed to the
1024 | memory system by CPU 1
David Howells108b42b2006-03-31 16:00:29 +01001025 V
1026
1027
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001028Secondly, data dependency barriers act as partial orderings on data-dependent
David Howells108b42b2006-03-31 16:00:29 +01001029loads. Consider the following sequence of events:
1030
1031 CPU 1 CPU 2
1032 ======================= =======================
David Howellsc14038c2006-04-10 22:54:24 -07001033 { B = 7; X = 9; Y = 8; C = &Y }
David Howells108b42b2006-03-31 16:00:29 +01001034 STORE A = 1
1035 STORE B = 2
1036 <write barrier>
1037 STORE C = &B LOAD X
1038 STORE D = 4 LOAD C (gets &B)
1039 LOAD *C (reads B)
1040
1041Without intervention, CPU 2 may perceive the events on CPU 1 in some
1042effectively random order, despite the write barrier issued by CPU 1:
1043
1044 +-------+ : : : :
1045 | | +------+ +-------+ | Sequence of update
1046 | |------>| B=2 |----- --->| Y->8 | | of perception on
1047 | | : +------+ \ +-------+ | CPU 2
1048 | CPU 1 | : | A=1 | \ --->| C->&Y | V
1049 | | +------+ | +-------+
1050 | | wwwwwwwwwwwwwwww | : :
1051 | | +------+ | : :
1052 | | : | C=&B |--- | : : +-------+
1053 | | : +------+ \ | +-------+ | |
1054 | |------>| D=4 | ----------->| C->&B |------>| |
1055 | | +------+ | +-------+ | |
1056 +-------+ : : | : : | |
1057 | : : | |
1058 | : : | CPU 2 |
1059 | +-------+ | |
1060 Apparently incorrect ---> | | B->7 |------>| |
1061 perception of B (!) | +-------+ | |
1062 | : : | |
1063 | +-------+ | |
1064 The load of X holds ---> \ | X->9 |------>| |
1065 up the maintenance \ +-------+ | |
1066 of coherence of B ----->| B->2 | +-------+
1067 +-------+
1068 : :
1069
1070
1071In the above example, CPU 2 perceives that B is 7, despite the load of *C
Paolo Ornati670e9f32006-10-03 22:57:56 +02001072(which would be B) coming after the LOAD of C.
David Howells108b42b2006-03-31 16:00:29 +01001073
1074If, however, a data dependency barrier were to be placed between the load of C
David Howellsc14038c2006-04-10 22:54:24 -07001075and the load of *C (ie: B) on CPU 2:
1076
1077 CPU 1 CPU 2
1078 ======================= =======================
1079 { B = 7; X = 9; Y = 8; C = &Y }
1080 STORE A = 1
1081 STORE B = 2
1082 <write barrier>
1083 STORE C = &B LOAD X
1084 STORE D = 4 LOAD C (gets &B)
1085 <data dependency barrier>
1086 LOAD *C (reads B)
1087
1088then the following will occur:
David Howells108b42b2006-03-31 16:00:29 +01001089
1090 +-------+ : : : :
1091 | | +------+ +-------+
1092 | |------>| B=2 |----- --->| Y->8 |
1093 | | : +------+ \ +-------+
1094 | CPU 1 | : | A=1 | \ --->| C->&Y |
1095 | | +------+ | +-------+
1096 | | wwwwwwwwwwwwwwww | : :
1097 | | +------+ | : :
1098 | | : | C=&B |--- | : : +-------+
1099 | | : +------+ \ | +-------+ | |
1100 | |------>| D=4 | ----------->| C->&B |------>| |
1101 | | +------+ | +-------+ | |
1102 +-------+ : : | : : | |
1103 | : : | |
1104 | : : | CPU 2 |
1105 | +-------+ | |
David Howells670bd952006-06-10 09:54:12 -07001106 | | X->9 |------>| |
1107 | +-------+ | |
1108 Makes sure all effects ---> \ ddddddddddddddddd | |
1109 prior to the store of C \ +-------+ | |
1110 are perceptible to ----->| B->2 |------>| |
1111 subsequent loads +-------+ | |
David Howells108b42b2006-03-31 16:00:29 +01001112 : : +-------+
1113
1114
1115And thirdly, a read barrier acts as a partial order on loads. Consider the
1116following sequence of events:
1117
1118 CPU 1 CPU 2
1119 ======================= =======================
David Howells670bd952006-06-10 09:54:12 -07001120 { A = 0, B = 9 }
David Howells108b42b2006-03-31 16:00:29 +01001121 STORE A=1
David Howells108b42b2006-03-31 16:00:29 +01001122 <write barrier>
David Howells670bd952006-06-10 09:54:12 -07001123 STORE B=2
David Howells108b42b2006-03-31 16:00:29 +01001124 LOAD B
David Howells670bd952006-06-10 09:54:12 -07001125 LOAD A
David Howells108b42b2006-03-31 16:00:29 +01001126
1127Without intervention, CPU 2 may then choose to perceive the events on CPU 1 in
1128some effectively random order, despite the write barrier issued by CPU 1:
1129
David Howells670bd952006-06-10 09:54:12 -07001130 +-------+ : : : :
1131 | | +------+ +-------+
1132 | |------>| A=1 |------ --->| A->0 |
1133 | | +------+ \ +-------+
1134 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1135 | | +------+ | +-------+
1136 | |------>| B=2 |--- | : :
1137 | | +------+ \ | : : +-------+
1138 +-------+ : : \ | +-------+ | |
1139 ---------->| B->2 |------>| |
1140 | +-------+ | CPU 2 |
1141 | | A->0 |------>| |
1142 | +-------+ | |
1143 | : : +-------+
1144 \ : :
1145 \ +-------+
1146 ---->| A->1 |
1147 +-------+
1148 : :
David Howells108b42b2006-03-31 16:00:29 +01001149
1150
David Howells6bc39272006-06-25 05:49:22 -07001151If, however, a read barrier were to be placed between the load of B and the
David Howells670bd952006-06-10 09:54:12 -07001152load of A on CPU 2:
David Howells108b42b2006-03-31 16:00:29 +01001153
David Howells670bd952006-06-10 09:54:12 -07001154 CPU 1 CPU 2
1155 ======================= =======================
1156 { A = 0, B = 9 }
1157 STORE A=1
1158 <write barrier>
1159 STORE B=2
1160 LOAD B
1161 <read barrier>
1162 LOAD A
1163
1164then the partial ordering imposed by CPU 1 will be perceived correctly by CPU
11652:
1166
1167 +-------+ : : : :
1168 | | +------+ +-------+
1169 | |------>| A=1 |------ --->| A->0 |
1170 | | +------+ \ +-------+
1171 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1172 | | +------+ | +-------+
1173 | |------>| B=2 |--- | : :
1174 | | +------+ \ | : : +-------+
1175 +-------+ : : \ | +-------+ | |
1176 ---------->| B->2 |------>| |
1177 | +-------+ | CPU 2 |
1178 | : : | |
1179 | : : | |
1180 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1181 barrier causes all effects \ +-------+ | |
1182 prior to the storage of B ---->| A->1 |------>| |
1183 to be perceptible to CPU 2 +-------+ | |
1184 : : +-------+
1185
1186
1187To illustrate this more completely, consider what could happen if the code
1188contained a load of A either side of the read barrier:
1189
1190 CPU 1 CPU 2
1191 ======================= =======================
1192 { A = 0, B = 9 }
1193 STORE A=1
1194 <write barrier>
1195 STORE B=2
1196 LOAD B
1197 LOAD A [first load of A]
1198 <read barrier>
1199 LOAD A [second load of A]
1200
1201Even though the two loads of A both occur after the load of B, they may both
1202come up with different values:
1203
1204 +-------+ : : : :
1205 | | +------+ +-------+
1206 | |------>| A=1 |------ --->| A->0 |
1207 | | +------+ \ +-------+
1208 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1209 | | +------+ | +-------+
1210 | |------>| B=2 |--- | : :
1211 | | +------+ \ | : : +-------+
1212 +-------+ : : \ | +-------+ | |
1213 ---------->| B->2 |------>| |
1214 | +-------+ | CPU 2 |
1215 | : : | |
1216 | : : | |
1217 | +-------+ | |
1218 | | A->0 |------>| 1st |
1219 | +-------+ | |
1220 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1221 barrier causes all effects \ +-------+ | |
1222 prior to the storage of B ---->| A->1 |------>| 2nd |
1223 to be perceptible to CPU 2 +-------+ | |
1224 : : +-------+
1225
1226
1227But it may be that the update to A from CPU 1 becomes perceptible to CPU 2
1228before the read barrier completes anyway:
1229
1230 +-------+ : : : :
1231 | | +------+ +-------+
1232 | |------>| A=1 |------ --->| A->0 |
1233 | | +------+ \ +-------+
1234 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1235 | | +------+ | +-------+
1236 | |------>| B=2 |--- | : :
1237 | | +------+ \ | : : +-------+
1238 +-------+ : : \ | +-------+ | |
1239 ---------->| B->2 |------>| |
1240 | +-------+ | CPU 2 |
1241 | : : | |
1242 \ : : | |
1243 \ +-------+ | |
1244 ---->| A->1 |------>| 1st |
1245 +-------+ | |
1246 rrrrrrrrrrrrrrrrr | |
1247 +-------+ | |
1248 | A->1 |------>| 2nd |
1249 +-------+ | |
1250 : : +-------+
1251
1252
1253The guarantee is that the second load will always come up with A == 1 if the
1254load of B came up with B == 2. No such guarantee exists for the first load of
1255A; that may come up with either A == 0 or A == 1.
1256
1257
1258READ MEMORY BARRIERS VS LOAD SPECULATION
1259----------------------------------------
1260
1261Many CPUs speculate with loads: that is they see that they will need to load an
1262item from memory, and they find a time where they're not using the bus for any
1263other loads, and so do the load in advance - even though they haven't actually
1264got to that point in the instruction execution flow yet. This permits the
1265actual load instruction to potentially complete immediately because the CPU
1266already has the value to hand.
1267
1268It may turn out that the CPU didn't actually need the value - perhaps because a
1269branch circumvented the load - in which case it can discard the value or just
1270cache it for later use.
1271
1272Consider:
1273
Ingo Molnare0edc782013-11-22 11:24:53 +01001274 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001275 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001276 LOAD B
1277 DIVIDE } Divide instructions generally
1278 DIVIDE } take a long time to perform
1279 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001280
1281Which might appear as this:
1282
1283 : : +-------+
1284 +-------+ | |
1285 --->| B->2 |------>| |
1286 +-------+ | CPU 2 |
1287 : :DIVIDE | |
1288 +-------+ | |
1289 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1290 division speculates on the +-------+ ~ | |
1291 LOAD of A : : ~ | |
1292 : :DIVIDE | |
1293 : : ~ | |
1294 Once the divisions are complete --> : : ~-->| |
1295 the CPU can then perform the : : | |
1296 LOAD with immediate effect : : +-------+
1297
1298
1299Placing a read barrier or a data dependency barrier just before the second
1300load:
1301
Ingo Molnare0edc782013-11-22 11:24:53 +01001302 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001303 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001304 LOAD B
1305 DIVIDE
1306 DIVIDE
David Howells670bd952006-06-10 09:54:12 -07001307 <read barrier>
Ingo Molnare0edc782013-11-22 11:24:53 +01001308 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001309
1310will force any value speculatively obtained to be reconsidered to an extent
1311dependent on the type of barrier used. If there was no change made to the
1312speculated memory location, then the speculated value will just be used:
1313
1314 : : +-------+
1315 +-------+ | |
1316 --->| B->2 |------>| |
1317 +-------+ | CPU 2 |
1318 : :DIVIDE | |
1319 +-------+ | |
1320 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1321 division speculates on the +-------+ ~ | |
1322 LOAD of A : : ~ | |
1323 : :DIVIDE | |
1324 : : ~ | |
1325 : : ~ | |
1326 rrrrrrrrrrrrrrrr~ | |
1327 : : ~ | |
1328 : : ~-->| |
1329 : : | |
1330 : : +-------+
1331
1332
1333but if there was an update or an invalidation from another CPU pending, then
1334the speculation will be cancelled and the value reloaded:
1335
1336 : : +-------+
1337 +-------+ | |
1338 --->| B->2 |------>| |
1339 +-------+ | CPU 2 |
1340 : :DIVIDE | |
1341 +-------+ | |
1342 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1343 division speculates on the +-------+ ~ | |
1344 LOAD of A : : ~ | |
1345 : :DIVIDE | |
1346 : : ~ | |
1347 : : ~ | |
1348 rrrrrrrrrrrrrrrrr | |
1349 +-------+ | |
1350 The speculation is discarded ---> --->| A->1 |------>| |
1351 and an updated value is +-------+ | |
1352 retrieved : : +-------+
David Howells108b42b2006-03-31 16:00:29 +01001353
1354
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001355MULTICOPY ATOMICITY
1356--------------------
Paul E. McKenney241e6662011-02-10 16:54:50 -08001357
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001358Multicopy atomicity is a deeply intuitive notion about ordering that is
1359not always provided by real computer systems, namely that a given store
Alan Stern0902b1f2017-09-01 07:53:34 -07001360becomes visible at the same time to all CPUs, or, alternatively, that all
1361CPUs agree on the order in which all stores become visible. However,
1362support of full multicopy atomicity would rule out valuable hardware
1363optimizations, so a weaker form called ``other multicopy atomicity''
1364instead guarantees only that a given store becomes visible at the same
1365time to all -other- CPUs. The remainder of this document discusses this
1366weaker form, but for brevity will call it simply ``multicopy atomicity''.
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001367
1368The following example demonstrates multicopy atomicity:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001369
1370 CPU 1 CPU 2 CPU 3
1371 ======================= ======================= =======================
1372 { X = 0, Y = 0 }
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001373 STORE X=1 r1=LOAD X (reads 1) LOAD Y (reads 1)
1374 <general barrier> <read barrier>
1375 STORE Y=r1 LOAD X
Paul E. McKenney241e6662011-02-10 16:54:50 -08001376
Alan Stern0902b1f2017-09-01 07:53:34 -07001377Suppose that CPU 2's load from X returns 1, which it then stores to Y,
1378and CPU 3's load from Y returns 1. This indicates that CPU 1's store
1379to X precedes CPU 2's load from X and that CPU 2's store to Y precedes
1380CPU 3's load from Y. In addition, the memory barriers guarantee that
1381CPU 2 executes its load before its store, and CPU 3 loads from Y before
1382it loads from X. The question is then "Can CPU 3's load from X return 0?"
Paul E. McKenney241e6662011-02-10 16:54:50 -08001383
Alan Stern0902b1f2017-09-01 07:53:34 -07001384Because CPU 3's load from X in some sense comes after CPU 2's load, it
Paul E. McKenney241e6662011-02-10 16:54:50 -08001385is natural to expect that CPU 3's load from X must therefore return 1.
Alan Stern0902b1f2017-09-01 07:53:34 -07001386This expectation follows from multicopy atomicity: if a load executing
1387on CPU B follows a load from the same variable executing on CPU A (and
1388CPU A did not originally store the value which it read), then on
1389multicopy-atomic systems, CPU B's load must return either the same value
1390that CPU A's load did or some later value. However, the Linux kernel
1391does not require systems to be multicopy atomic.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001392
Alan Stern0902b1f2017-09-01 07:53:34 -07001393The use of a general memory barrier in the example above compensates
1394for any lack of multicopy atomicity. In the example, if CPU 2's load
1395from X returns 1 and CPU 3's load from Y returns 1, then CPU 3's load
1396from X must indeed also return 1.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001397
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001398However, dependencies, read barriers, and write barriers are not always
1399able to compensate for non-multicopy atomicity. For example, suppose
1400that CPU 2's general barrier is removed from the above example, leaving
1401only the data dependency shown below:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001402
1403 CPU 1 CPU 2 CPU 3
1404 ======================= ======================= =======================
1405 { X = 0, Y = 0 }
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001406 STORE X=1 r1=LOAD X (reads 1) LOAD Y (reads 1)
1407 <data dependency> <read barrier>
1408 STORE Y=r1 LOAD X (reads 0)
Paul E. McKenney241e6662011-02-10 16:54:50 -08001409
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001410This substitution allows non-multicopy atomicity to run rampant: in
1411this example, it is perfectly legal for CPU 2's load from X to return 1,
1412CPU 3's load from Y to return 1, and its load from X to return 0.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001413
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001414The key point is that although CPU 2's data dependency orders its load
Alan Stern0902b1f2017-09-01 07:53:34 -07001415and store, it does not guarantee to order CPU 1's store. Thus, if this
1416example runs on a non-multicopy-atomic system where CPUs 1 and 2 share a
1417store buffer or a level of cache, CPU 2 might have early access to CPU 1's
1418writes. General barriers are therefore required to ensure that all CPUs
1419agree on the combined order of multiple accesses.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001420
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001421General barriers can compensate not only for non-multicopy atomicity,
1422but can also generate additional ordering that can ensure that -all-
1423CPUs will perceive the same order of -all- operations. In contrast, a
1424chain of release-acquire pairs do not provide this additional ordering,
1425which means that only those CPUs on the chain are guaranteed to agree
1426on the combined order of the accesses. For example, switching to C code
1427in deference to the ghost of Herman Hollerith:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001428
1429 int u, v, x, y, z;
1430
1431 void cpu0(void)
1432 {
1433 r0 = smp_load_acquire(&x);
1434 WRITE_ONCE(u, 1);
1435 smp_store_release(&y, 1);
1436 }
1437
1438 void cpu1(void)
1439 {
1440 r1 = smp_load_acquire(&y);
1441 r4 = READ_ONCE(v);
1442 r5 = READ_ONCE(u);
1443 smp_store_release(&z, 1);
1444 }
1445
1446 void cpu2(void)
1447 {
1448 r2 = smp_load_acquire(&z);
1449 smp_store_release(&x, 1);
1450 }
1451
1452 void cpu3(void)
1453 {
1454 WRITE_ONCE(v, 1);
1455 smp_mb();
1456 r3 = READ_ONCE(u);
1457 }
1458
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001459Because cpu0(), cpu1(), and cpu2() participate in a chain of
1460smp_store_release()/smp_load_acquire() pairs, the following outcome
1461is prohibited:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001462
1463 r0 == 1 && r1 == 1 && r2 == 1
1464
1465Furthermore, because of the release-acquire relationship between cpu0()
1466and cpu1(), cpu1() must see cpu0()'s writes, so that the following
1467outcome is prohibited:
1468
1469 r1 == 1 && r5 == 0
1470
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001471However, the ordering provided by a release-acquire chain is local
1472to the CPUs participating in that chain and does not apply to cpu3(),
1473at least aside from stores. Therefore, the following outcome is possible:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001474
1475 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0
1476
Paul E. McKenney37ef0342016-01-25 22:12:34 -08001477As an aside, the following outcome is also possible:
1478
1479 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0 && r5 == 1
1480
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001481Although cpu0(), cpu1(), and cpu2() will see their respective reads and
1482writes in order, CPUs not involved in the release-acquire chain might
1483well disagree on the order. This disagreement stems from the fact that
1484the weak memory-barrier instructions used to implement smp_load_acquire()
1485and smp_store_release() are not required to order prior stores against
1486subsequent loads in all cases. This means that cpu3() can see cpu0()'s
1487store to u as happening -after- cpu1()'s load from v, even though
1488both cpu0() and cpu1() agree that these two operations occurred in the
1489intended order.
1490
1491However, please keep in mind that smp_load_acquire() is not magic.
1492In particular, it simply reads from its argument with ordering. It does
1493-not- ensure that any particular value will be read. Therefore, the
1494following outcome is possible:
1495
1496 r0 == 0 && r1 == 0 && r2 == 0 && r5 == 0
1497
1498Note that this outcome can happen even on a mythical sequentially
1499consistent system where nothing is ever reordered.
1500
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001501To reiterate, if your code requires full ordering of all operations,
1502use general barriers throughout.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001503
1504
David Howells108b42b2006-03-31 16:00:29 +01001505========================
1506EXPLICIT KERNEL BARRIERS
1507========================
1508
1509The Linux kernel has a variety of different barriers that act at different
1510levels:
1511
1512 (*) Compiler barrier.
1513
1514 (*) CPU memory barriers.
1515
1516 (*) MMIO write barrier.
1517
1518
1519COMPILER BARRIER
1520----------------
1521
1522The Linux kernel has an explicit compiler barrier function that prevents the
1523compiler from moving the memory accesses either side of it to the other side:
1524
1525 barrier();
1526
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001527This is a general barrier -- there are no read-read or write-write
1528variants of barrier(). However, READ_ONCE() and WRITE_ONCE() can be
1529thought of as weak forms of barrier() that affect only the specific
1530accesses flagged by the READ_ONCE() or WRITE_ONCE().
David Howells108b42b2006-03-31 16:00:29 +01001531
Paul E. McKenney692118d2013-12-11 13:59:07 -08001532The barrier() function has the following effects:
1533
1534 (*) Prevents the compiler from reordering accesses following the
1535 barrier() to precede any accesses preceding the barrier().
1536 One example use for this property is to ease communication between
1537 interrupt-handler code and the code that was interrupted.
1538
1539 (*) Within a loop, forces the compiler to load the variables used
1540 in that loop's conditional on each pass through that loop.
1541
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001542The READ_ONCE() and WRITE_ONCE() functions can prevent any number of
1543optimizations that, while perfectly safe in single-threaded code, can
1544be fatal in concurrent code. Here are some examples of these sorts
1545of optimizations:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001546
Paul E. McKenney449f7412014-01-02 15:03:50 -08001547 (*) The compiler is within its rights to reorder loads and stores
1548 to the same variable, and in some cases, the CPU is within its
1549 rights to reorder loads to the same variable. This means that
1550 the following code:
1551
1552 a[0] = x;
1553 a[1] = x;
1554
1555 Might result in an older value of x stored in a[1] than in a[0].
1556 Prevent both the compiler and the CPU from doing this as follows:
1557
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001558 a[0] = READ_ONCE(x);
1559 a[1] = READ_ONCE(x);
Paul E. McKenney449f7412014-01-02 15:03:50 -08001560
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001561 In short, READ_ONCE() and WRITE_ONCE() provide cache coherence for
1562 accesses from multiple CPUs to a single variable.
Paul E. McKenney449f7412014-01-02 15:03:50 -08001563
Paul E. McKenney692118d2013-12-11 13:59:07 -08001564 (*) The compiler is within its rights to merge successive loads from
1565 the same variable. Such merging can cause the compiler to "optimize"
1566 the following code:
1567
1568 while (tmp = a)
1569 do_something_with(tmp);
1570
1571 into the following code, which, although in some sense legitimate
1572 for single-threaded code, is almost certainly not what the developer
1573 intended:
1574
1575 if (tmp = a)
1576 for (;;)
1577 do_something_with(tmp);
1578
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001579 Use READ_ONCE() to prevent the compiler from doing this to you:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001580
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001581 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001582 do_something_with(tmp);
1583
1584 (*) The compiler is within its rights to reload a variable, for example,
1585 in cases where high register pressure prevents the compiler from
1586 keeping all data of interest in registers. The compiler might
1587 therefore optimize the variable 'tmp' out of our previous example:
1588
1589 while (tmp = a)
1590 do_something_with(tmp);
1591
1592 This could result in the following code, which is perfectly safe in
1593 single-threaded code, but can be fatal in concurrent code:
1594
1595 while (a)
1596 do_something_with(a);
1597
1598 For example, the optimized version of this code could result in
1599 passing a zero to do_something_with() in the case where the variable
1600 a was modified by some other CPU between the "while" statement and
1601 the call to do_something_with().
1602
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001603 Again, use READ_ONCE() to prevent the compiler from doing this:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001604
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001605 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001606 do_something_with(tmp);
1607
1608 Note that if the compiler runs short of registers, it might save
1609 tmp onto the stack. The overhead of this saving and later restoring
1610 is why compilers reload variables. Doing so is perfectly safe for
1611 single-threaded code, so you need to tell the compiler about cases
1612 where it is not safe.
1613
1614 (*) The compiler is within its rights to omit a load entirely if it knows
1615 what the value will be. For example, if the compiler can prove that
1616 the value of variable 'a' is always zero, it can optimize this code:
1617
1618 while (tmp = a)
1619 do_something_with(tmp);
1620
1621 Into this:
1622
1623 do { } while (0);
1624
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001625 This transformation is a win for single-threaded code because it
1626 gets rid of a load and a branch. The problem is that the compiler
1627 will carry out its proof assuming that the current CPU is the only
1628 one updating variable 'a'. If variable 'a' is shared, then the
1629 compiler's proof will be erroneous. Use READ_ONCE() to tell the
1630 compiler that it doesn't know as much as it thinks it does:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001631
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001632 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001633 do_something_with(tmp);
1634
1635 But please note that the compiler is also closely watching what you
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001636 do with the value after the READ_ONCE(). For example, suppose you
Paul E. McKenney692118d2013-12-11 13:59:07 -08001637 do the following and MAX is a preprocessor macro with the value 1:
1638
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001639 while ((tmp = READ_ONCE(a)) % MAX)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001640 do_something_with(tmp);
1641
1642 Then the compiler knows that the result of the "%" operator applied
1643 to MAX will always be zero, again allowing the compiler to optimize
1644 the code into near-nonexistence. (It will still load from the
1645 variable 'a'.)
1646
1647 (*) Similarly, the compiler is within its rights to omit a store entirely
1648 if it knows that the variable already has the value being stored.
1649 Again, the compiler assumes that the current CPU is the only one
1650 storing into the variable, which can cause the compiler to do the
1651 wrong thing for shared variables. For example, suppose you have
1652 the following:
1653
1654 a = 0;
SeongJae Park65f95ff2016-02-22 08:28:29 -08001655 ... Code that does not store to variable a ...
Paul E. McKenney692118d2013-12-11 13:59:07 -08001656 a = 0;
1657
1658 The compiler sees that the value of variable 'a' is already zero, so
1659 it might well omit the second store. This would come as a fatal
1660 surprise if some other CPU might have stored to variable 'a' in the
1661 meantime.
1662
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001663 Use WRITE_ONCE() to prevent the compiler from making this sort of
Paul E. McKenney692118d2013-12-11 13:59:07 -08001664 wrong guess:
1665
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001666 WRITE_ONCE(a, 0);
SeongJae Park65f95ff2016-02-22 08:28:29 -08001667 ... Code that does not store to variable a ...
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001668 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001669
1670 (*) The compiler is within its rights to reorder memory accesses unless
1671 you tell it not to. For example, consider the following interaction
1672 between process-level code and an interrupt handler:
1673
1674 void process_level(void)
1675 {
1676 msg = get_message();
1677 flag = true;
1678 }
1679
1680 void interrupt_handler(void)
1681 {
1682 if (flag)
1683 process_message(msg);
1684 }
1685
Masanari Iidadf5cbb22014-03-21 10:04:30 +09001686 There is nothing to prevent the compiler from transforming
Paul E. McKenney692118d2013-12-11 13:59:07 -08001687 process_level() to the following, in fact, this might well be a
1688 win for single-threaded code:
1689
1690 void process_level(void)
1691 {
1692 flag = true;
1693 msg = get_message();
1694 }
1695
1696 If the interrupt occurs between these two statement, then
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001697 interrupt_handler() might be passed a garbled msg. Use WRITE_ONCE()
Paul E. McKenney692118d2013-12-11 13:59:07 -08001698 to prevent this as follows:
1699
1700 void process_level(void)
1701 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001702 WRITE_ONCE(msg, get_message());
1703 WRITE_ONCE(flag, true);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001704 }
1705
1706 void interrupt_handler(void)
1707 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001708 if (READ_ONCE(flag))
1709 process_message(READ_ONCE(msg));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001710 }
1711
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001712 Note that the READ_ONCE() and WRITE_ONCE() wrappers in
1713 interrupt_handler() are needed if this interrupt handler can itself
1714 be interrupted by something that also accesses 'flag' and 'msg',
1715 for example, a nested interrupt or an NMI. Otherwise, READ_ONCE()
1716 and WRITE_ONCE() are not needed in interrupt_handler() other than
1717 for documentation purposes. (Note also that nested interrupts
1718 do not typically occur in modern Linux kernels, in fact, if an
1719 interrupt handler returns with interrupts enabled, you will get a
1720 WARN_ONCE() splat.)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001721
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001722 You should assume that the compiler can move READ_ONCE() and
1723 WRITE_ONCE() past code not containing READ_ONCE(), WRITE_ONCE(),
1724 barrier(), or similar primitives.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001725
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001726 This effect could also be achieved using barrier(), but READ_ONCE()
1727 and WRITE_ONCE() are more selective: With READ_ONCE() and
1728 WRITE_ONCE(), the compiler need only forget the contents of the
1729 indicated memory locations, while with barrier() the compiler must
1730 discard the value of all memory locations that it has currented
1731 cached in any machine registers. Of course, the compiler must also
1732 respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
1733 though the CPU of course need not do so.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001734
1735 (*) The compiler is within its rights to invent stores to a variable,
1736 as in the following example:
1737
1738 if (a)
1739 b = a;
1740 else
1741 b = 42;
1742
1743 The compiler might save a branch by optimizing this as follows:
1744
1745 b = 42;
1746 if (a)
1747 b = a;
1748
1749 In single-threaded code, this is not only safe, but also saves
1750 a branch. Unfortunately, in concurrent code, this optimization
1751 could cause some other CPU to see a spurious value of 42 -- even
1752 if variable 'a' was never zero -- when loading variable 'b'.
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001753 Use WRITE_ONCE() to prevent this as follows:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001754
1755 if (a)
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001756 WRITE_ONCE(b, a);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001757 else
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001758 WRITE_ONCE(b, 42);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001759
1760 The compiler can also invent loads. These are usually less
1761 damaging, but they can result in cache-line bouncing and thus in
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001762 poor performance and scalability. Use READ_ONCE() to prevent
Paul E. McKenney692118d2013-12-11 13:59:07 -08001763 invented loads.
1764
1765 (*) For aligned memory locations whose size allows them to be accessed
1766 with a single memory-reference instruction, prevents "load tearing"
1767 and "store tearing," in which a single large access is replaced by
1768 multiple smaller accesses. For example, given an architecture having
1769 16-bit store instructions with 7-bit immediate fields, the compiler
1770 might be tempted to use two 16-bit store-immediate instructions to
1771 implement the following 32-bit store:
1772
1773 p = 0x00010002;
1774
1775 Please note that GCC really does use this sort of optimization,
1776 which is not surprising given that it would likely take more
1777 than two instructions to build the constant and then store it.
1778 This optimization can therefore be a win in single-threaded code.
1779 In fact, a recent bug (since fixed) caused GCC to incorrectly use
1780 this optimization in a volatile store. In the absence of such bugs,
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001781 use of WRITE_ONCE() prevents store tearing in the following example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001782
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001783 WRITE_ONCE(p, 0x00010002);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001784
1785 Use of packed structures can also result in load and store tearing,
1786 as in this example:
1787
1788 struct __attribute__((__packed__)) foo {
1789 short a;
1790 int b;
1791 short c;
1792 };
1793 struct foo foo1, foo2;
1794 ...
1795
1796 foo2.a = foo1.a;
1797 foo2.b = foo1.b;
1798 foo2.c = foo1.c;
1799
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001800 Because there are no READ_ONCE() or WRITE_ONCE() wrappers and no
1801 volatile markings, the compiler would be well within its rights to
1802 implement these three assignment statements as a pair of 32-bit
1803 loads followed by a pair of 32-bit stores. This would result in
1804 load tearing on 'foo1.b' and store tearing on 'foo2.b'. READ_ONCE()
1805 and WRITE_ONCE() again prevent tearing in this example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001806
1807 foo2.a = foo1.a;
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001808 WRITE_ONCE(foo2.b, READ_ONCE(foo1.b));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001809 foo2.c = foo1.c;
1810
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001811All that aside, it is never necessary to use READ_ONCE() and
1812WRITE_ONCE() on a variable that has been marked volatile. For example,
1813because 'jiffies' is marked volatile, it is never necessary to
1814say READ_ONCE(jiffies). The reason for this is that READ_ONCE() and
1815WRITE_ONCE() are implemented as volatile casts, which has no effect when
1816its argument is already marked volatile.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001817
1818Please note that these compiler barriers have no direct effect on the CPU,
1819which may then reorder things however it wishes.
David Howells108b42b2006-03-31 16:00:29 +01001820
1821
1822CPU MEMORY BARRIERS
1823-------------------
1824
1825The Linux kernel has eight basic CPU memory barriers:
1826
1827 TYPE MANDATORY SMP CONDITIONAL
1828 =============== ======================= ===========================
1829 GENERAL mb() smp_mb()
1830 WRITE wmb() smp_wmb()
1831 READ rmb() smp_rmb()
Paul E. McKenney9ad3c142017-11-27 09:20:40 -08001832 DATA DEPENDENCY READ_ONCE()
David Howells108b42b2006-03-31 16:00:29 +01001833
1834
Nick Piggin73f10282008-05-14 06:35:11 +02001835All memory barriers except the data dependency barriers imply a compiler
SeongJae Park0b6fa342016-04-12 08:52:53 -07001836barrier. Data dependencies do not impose any additional compiler ordering.
Nick Piggin73f10282008-05-14 06:35:11 +02001837
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001838Aside: In the case of data dependencies, the compiler would be expected
1839to issue the loads in the correct order (eg. `a[b]` would have to load
1840the value of b before loading a[b]), however there is no guarantee in
1841the C specification that the compiler may not speculate the value of b
1842(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
SeongJae Park0b6fa342016-04-12 08:52:53 -07001843tmp = a[b]; ). There is also the problem of a compiler reloading b after
1844having loaded a[b], thus having a newer copy of b than a[b]. A consensus
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001845has not yet been reached about these problems, however the READ_ONCE()
1846macro is a good place to start looking.
David Howells108b42b2006-03-31 16:00:29 +01001847
1848SMP memory barriers are reduced to compiler barriers on uniprocessor compiled
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001849systems because it is assumed that a CPU will appear to be self-consistent,
David Howells108b42b2006-03-31 16:00:29 +01001850and will order overlapping accesses correctly with respect to itself.
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001851However, see the subsection on "Virtual Machine Guests" below.
David Howells108b42b2006-03-31 16:00:29 +01001852
1853[!] Note that SMP memory barriers _must_ be used to control the ordering of
1854references to shared memory on SMP systems, though the use of locking instead
1855is sufficient.
1856
1857Mandatory barriers should not be used to control SMP effects, since mandatory
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001858barriers impose unnecessary overhead on both SMP and UP systems. They may,
1859however, be used to control MMIO effects on accesses through relaxed memory I/O
1860windows. These barriers are required even on non-SMP systems as they affect
1861the order in which memory operations appear to a device by prohibiting both the
1862compiler and the CPU from reordering them.
David Howells108b42b2006-03-31 16:00:29 +01001863
1864
1865There are some more advanced barrier functions:
1866
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001867 (*) smp_store_mb(var, value)
David Howells108b42b2006-03-31 16:00:29 +01001868
Oleg Nesterov75b2bd52006-11-08 17:44:38 -08001869 This assigns the value to the variable and then inserts a full memory
Davidlohr Bueso2d142e52015-10-27 12:53:51 -07001870 barrier after it. It isn't guaranteed to insert anything more than a
1871 compiler barrier in a UP compilation.
David Howells108b42b2006-03-31 16:00:29 +01001872
1873
Peter Zijlstra1b156112014-03-13 19:00:35 +01001874 (*) smp_mb__before_atomic();
1875 (*) smp_mb__after_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001876
Peter Zijlstra1b156112014-03-13 19:00:35 +01001877 These are for use with atomic (such as add, subtract, increment and
1878 decrement) functions that don't return a value, especially when used for
1879 reference counting. These functions do not imply memory barriers.
1880
1881 These are also used for atomic bitop functions that do not return a
1882 value (such as set_bit and clear_bit).
David Howells108b42b2006-03-31 16:00:29 +01001883
1884 As an example, consider a piece of code that marks an object as being dead
1885 and then decrements the object's reference count:
1886
1887 obj->dead = 1;
Peter Zijlstra1b156112014-03-13 19:00:35 +01001888 smp_mb__before_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001889 atomic_dec(&obj->ref_count);
1890
1891 This makes sure that the death mark on the object is perceived to be set
1892 *before* the reference counter is decremented.
1893
Peter Zijlstra706eeb32017-06-12 14:50:27 +02001894 See Documentation/atomic_{t,bitops}.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01001895
1896
Alexander Duyck1077fa32014-12-11 15:02:06 -08001897 (*) dma_wmb();
1898 (*) dma_rmb();
1899
1900 These are for use with consistent memory to guarantee the ordering
1901 of writes or reads of shared memory accessible to both the CPU and a
1902 DMA capable device.
1903
1904 For example, consider a device driver that shares memory with a device
1905 and uses a descriptor status value to indicate if the descriptor belongs
1906 to the device or the CPU, and a doorbell to notify it when new
1907 descriptors are available:
1908
1909 if (desc->status != DEVICE_OWN) {
1910 /* do not read data until we own descriptor */
1911 dma_rmb();
1912
1913 /* read/modify data */
1914 read_data = desc->data;
1915 desc->data = write_data;
1916
1917 /* flush modifications before status update */
1918 dma_wmb();
1919
1920 /* assign ownership */
1921 desc->status = DEVICE_OWN;
1922
1923 /* force memory to sync before notifying device via MMIO */
1924 wmb();
1925
1926 /* notify device of new descriptors */
1927 writel(DESC_NOTIFY, doorbell);
1928 }
1929
1930 The dma_rmb() allows us guarantee the device has released ownership
Sylvain Trias7a458002015-04-08 10:27:57 +02001931 before we read the data from the descriptor, and the dma_wmb() allows
Alexander Duyck1077fa32014-12-11 15:02:06 -08001932 us to guarantee the data is written to the descriptor before the device
1933 can see it now has ownership. The wmb() is needed to guarantee that the
1934 cache coherent memory writes have completed before attempting a write to
1935 the cache incoherent MMIO region.
1936
1937 See Documentation/DMA-API.txt for more information on consistent memory.
1938
SeongJae Parkdfeccea2016-08-11 11:17:40 -07001939
David Howells108b42b2006-03-31 16:00:29 +01001940MMIO WRITE BARRIER
1941------------------
1942
1943The Linux kernel also has a special barrier for use with memory-mapped I/O
1944writes:
1945
1946 mmiowb();
1947
1948This is a variation on the mandatory write barrier that causes writes to weakly
1949ordered I/O regions to be partially ordered. Its effects may go beyond the
1950CPU->Hardware interface and actually affect the hardware at some level.
1951
SeongJae Park166bda72016-04-12 08:52:50 -07001952See the subsection "Acquires vs I/O accesses" for more information.
David Howells108b42b2006-03-31 16:00:29 +01001953
1954
1955===============================
1956IMPLICIT KERNEL MEMORY BARRIERS
1957===============================
1958
1959Some of the other functions in the linux kernel imply memory barriers, amongst
David Howells670bd952006-06-10 09:54:12 -07001960which are locking and scheduling functions.
David Howells108b42b2006-03-31 16:00:29 +01001961
1962This specification is a _minimum_ guarantee; any particular architecture may
1963provide more substantial guarantees, but these may not be relied upon outside
1964of arch specific code.
1965
1966
SeongJae Park166bda72016-04-12 08:52:50 -07001967LOCK ACQUISITION FUNCTIONS
1968--------------------------
David Howells108b42b2006-03-31 16:00:29 +01001969
1970The Linux kernel has a number of locking constructs:
1971
1972 (*) spin locks
1973 (*) R/W spin locks
1974 (*) mutexes
1975 (*) semaphores
1976 (*) R/W semaphores
David Howells108b42b2006-03-31 16:00:29 +01001977
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001978In all cases there are variants on "ACQUIRE" operations and "RELEASE" operations
David Howells108b42b2006-03-31 16:00:29 +01001979for each construct. These operations all imply certain barriers:
1980
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001981 (1) ACQUIRE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001982
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001983 Memory operations issued after the ACQUIRE will be completed after the
1984 ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001985
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001986 Memory operations issued before the ACQUIRE may be completed after
Peter Zijlstraa9668cd2017-06-07 17:51:27 +02001987 the ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001988
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001989 (2) RELEASE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001990
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001991 Memory operations issued before the RELEASE will be completed before the
1992 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001993
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001994 Memory operations issued after the RELEASE may be completed before the
1995 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001996
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001997 (3) ACQUIRE vs ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001998
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001999 All ACQUIRE operations issued before another ACQUIRE operation will be
2000 completed before that ACQUIRE operation.
David Howells108b42b2006-03-31 16:00:29 +01002001
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002002 (4) ACQUIRE vs RELEASE implication:
David Howells108b42b2006-03-31 16:00:29 +01002003
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002004 All ACQUIRE operations issued before a RELEASE operation will be
2005 completed before the RELEASE operation.
David Howells108b42b2006-03-31 16:00:29 +01002006
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002007 (5) Failed conditional ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01002008
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002009 Certain locking variants of the ACQUIRE operation may fail, either due to
2010 being unable to get the lock immediately, or due to receiving an unblocked
David Howells108b42b2006-03-31 16:00:29 +01002011 signal whilst asleep waiting for the lock to become available. Failed
2012 locks do not imply any sort of barrier.
2013
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002014[!] Note: one of the consequences of lock ACQUIREs and RELEASEs being only
2015one-way barriers is that the effects of instructions outside of a critical
2016section may seep into the inside of the critical section.
David Howells108b42b2006-03-31 16:00:29 +01002017
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002018An ACQUIRE followed by a RELEASE may not be assumed to be full memory barrier
2019because it is possible for an access preceding the ACQUIRE to happen after the
2020ACQUIRE, and an access following the RELEASE to happen before the RELEASE, and
2021the two accesses can themselves then cross:
David Howells670bd952006-06-10 09:54:12 -07002022
2023 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002024 ACQUIRE M
2025 RELEASE M
David Howells670bd952006-06-10 09:54:12 -07002026 *B = b;
2027
2028may occur as:
2029
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002030 ACQUIRE M, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002031
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002032When the ACQUIRE and RELEASE are a lock acquisition and release,
2033respectively, this same reordering can occur if the lock's ACQUIRE and
2034RELEASE are to the same lock variable, but only from the perspective of
2035another CPU not holding that lock. In short, a ACQUIRE followed by an
2036RELEASE may -not- be assumed to be a full memory barrier.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002037
Paul E. McKenney12d560f2015-07-14 18:35:23 -07002038Similarly, the reverse case of a RELEASE followed by an ACQUIRE does
2039not imply a full memory barrier. Therefore, the CPU's execution of the
2040critical sections corresponding to the RELEASE and the ACQUIRE can cross,
2041so that:
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002042
2043 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002044 RELEASE M
2045 ACQUIRE N
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002046 *B = b;
2047
2048could occur as:
2049
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002050 ACQUIRE N, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002051
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002052It might appear that this reordering could introduce a deadlock.
2053However, this cannot happen because if such a deadlock threatened,
2054the RELEASE would simply complete, thereby avoiding the deadlock.
2055
2056 Why does this work?
2057
2058 One key point is that we are only talking about the CPU doing
2059 the reordering, not the compiler. If the compiler (or, for
2060 that matter, the developer) switched the operations, deadlock
2061 -could- occur.
2062
2063 But suppose the CPU reordered the operations. In this case,
2064 the unlock precedes the lock in the assembly code. The CPU
2065 simply elected to try executing the later lock operation first.
2066 If there is a deadlock, this lock operation will simply spin (or
2067 try to sleep, but more on that later). The CPU will eventually
2068 execute the unlock operation (which preceded the lock operation
2069 in the assembly code), which will unravel the potential deadlock,
2070 allowing the lock operation to succeed.
2071
2072 But what if the lock is a sleeplock? In that case, the code will
2073 try to enter the scheduler, where it will eventually encounter
2074 a memory barrier, which will force the earlier unlock operation
2075 to complete, again unraveling the deadlock. There might be
2076 a sleep-unlock race, but the locking primitive needs to resolve
2077 such races properly in any case.
2078
David Howells108b42b2006-03-31 16:00:29 +01002079Locks and semaphores may not provide any guarantee of ordering on UP compiled
2080systems, and so cannot be counted on in such a situation to actually achieve
2081anything at all - especially with respect to I/O accesses - unless combined
2082with interrupt disabling operations.
2083
SeongJae Parkd7cab362016-08-11 11:17:41 -07002084See also the section on "Inter-CPU acquiring barrier effects".
David Howells108b42b2006-03-31 16:00:29 +01002085
2086
2087As an example, consider the following:
2088
2089 *A = a;
2090 *B = b;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002091 ACQUIRE
David Howells108b42b2006-03-31 16:00:29 +01002092 *C = c;
2093 *D = d;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002094 RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002095 *E = e;
2096 *F = f;
2097
2098The following sequence of events is acceptable:
2099
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002100 ACQUIRE, {*F,*A}, *E, {*C,*D}, *B, RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002101
2102 [+] Note that {*F,*A} indicates a combined access.
2103
2104But none of the following are:
2105
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002106 {*F,*A}, *B, ACQUIRE, *C, *D, RELEASE, *E
2107 *A, *B, *C, ACQUIRE, *D, RELEASE, *E, *F
2108 *A, *B, ACQUIRE, *C, RELEASE, *D, *E, *F
2109 *B, ACQUIRE, *C, *D, RELEASE, {*F,*A}, *E
David Howells108b42b2006-03-31 16:00:29 +01002110
2111
2112
2113INTERRUPT DISABLING FUNCTIONS
2114-----------------------------
2115
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002116Functions that disable interrupts (ACQUIRE equivalent) and enable interrupts
2117(RELEASE equivalent) will act as compiler barriers only. So if memory or I/O
David Howells108b42b2006-03-31 16:00:29 +01002118barriers are required in such a situation, they must be provided from some
2119other means.
2120
2121
David Howells50fa6102009-04-28 15:01:38 +01002122SLEEP AND WAKE-UP FUNCTIONS
2123---------------------------
2124
2125Sleeping and waking on an event flagged in global data can be viewed as an
2126interaction between two pieces of data: the task state of the task waiting for
2127the event and the global data used to indicate the event. To make sure that
2128these appear to happen in the right order, the primitives to begin the process
2129of going to sleep, and the primitives to initiate a wake up imply certain
2130barriers.
2131
2132Firstly, the sleeper normally follows something like this sequence of events:
2133
2134 for (;;) {
2135 set_current_state(TASK_UNINTERRUPTIBLE);
2136 if (event_indicated)
2137 break;
2138 schedule();
2139 }
2140
2141A general memory barrier is interpolated automatically by set_current_state()
2142after it has altered the task state:
2143
2144 CPU 1
2145 ===============================
2146 set_current_state();
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002147 smp_store_mb();
David Howells50fa6102009-04-28 15:01:38 +01002148 STORE current->state
2149 <general barrier>
2150 LOAD event_indicated
2151
2152set_current_state() may be wrapped by:
2153
2154 prepare_to_wait();
2155 prepare_to_wait_exclusive();
2156
2157which therefore also imply a general memory barrier after setting the state.
2158The whole sequence above is available in various canned forms, all of which
2159interpolate the memory barrier in the right place:
2160
2161 wait_event();
2162 wait_event_interruptible();
2163 wait_event_interruptible_exclusive();
2164 wait_event_interruptible_timeout();
2165 wait_event_killable();
2166 wait_event_timeout();
2167 wait_on_bit();
2168 wait_on_bit_lock();
2169
2170
2171Secondly, code that performs a wake up normally follows something like this:
2172
2173 event_indicated = 1;
2174 wake_up(&event_wait_queue);
2175
2176or:
2177
2178 event_indicated = 1;
2179 wake_up_process(event_daemon);
2180
SeongJae Park0b6fa342016-04-12 08:52:53 -07002181A write memory barrier is implied by wake_up() and co. if and only if they
2182wake something up. The barrier occurs before the task state is cleared, and so
2183sits between the STORE to indicate the event and the STORE to set TASK_RUNNING:
David Howells50fa6102009-04-28 15:01:38 +01002184
2185 CPU 1 CPU 2
2186 =============================== ===============================
2187 set_current_state(); STORE event_indicated
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002188 smp_store_mb(); wake_up();
David Howells50fa6102009-04-28 15:01:38 +01002189 STORE current->state <write barrier>
2190 <general barrier> STORE current->state
2191 LOAD event_indicated
2192
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002193To repeat, this write memory barrier is present if and only if something
2194is actually awakened. To see this, consider the following sequence of
2195events, where X and Y are both initially zero:
2196
2197 CPU 1 CPU 2
2198 =============================== ===============================
2199 X = 1; STORE event_indicated
2200 smp_mb(); wake_up();
2201 Y = 1; wait_event(wq, Y == 1);
2202 wake_up(); load from Y sees 1, no memory barrier
2203 load from X might see 0
2204
2205In contrast, if a wakeup does occur, CPU 2's load from X would be guaranteed
2206to see 1.
2207
David Howells50fa6102009-04-28 15:01:38 +01002208The available waker functions include:
2209
2210 complete();
2211 wake_up();
2212 wake_up_all();
2213 wake_up_bit();
2214 wake_up_interruptible();
2215 wake_up_interruptible_all();
2216 wake_up_interruptible_nr();
2217 wake_up_interruptible_poll();
2218 wake_up_interruptible_sync();
2219 wake_up_interruptible_sync_poll();
2220 wake_up_locked();
2221 wake_up_locked_poll();
2222 wake_up_nr();
2223 wake_up_poll();
2224 wake_up_process();
2225
2226
2227[!] Note that the memory barriers implied by the sleeper and the waker do _not_
2228order multiple stores before the wake-up with respect to loads of those stored
2229values after the sleeper has called set_current_state(). For instance, if the
2230sleeper does:
2231
2232 set_current_state(TASK_INTERRUPTIBLE);
2233 if (event_indicated)
2234 break;
2235 __set_current_state(TASK_RUNNING);
2236 do_something(my_data);
2237
2238and the waker does:
2239
2240 my_data = value;
2241 event_indicated = 1;
2242 wake_up(&event_wait_queue);
2243
2244there's no guarantee that the change to event_indicated will be perceived by
2245the sleeper as coming after the change to my_data. In such a circumstance, the
2246code on both sides must interpolate its own memory barriers between the
2247separate data accesses. Thus the above sleeper ought to do:
2248
2249 set_current_state(TASK_INTERRUPTIBLE);
2250 if (event_indicated) {
2251 smp_rmb();
2252 do_something(my_data);
2253 }
2254
2255and the waker should do:
2256
2257 my_data = value;
2258 smp_wmb();
2259 event_indicated = 1;
2260 wake_up(&event_wait_queue);
2261
2262
David Howells108b42b2006-03-31 16:00:29 +01002263MISCELLANEOUS FUNCTIONS
2264-----------------------
2265
2266Other functions that imply barriers:
2267
2268 (*) schedule() and similar imply full memory barriers.
2269
David Howells108b42b2006-03-31 16:00:29 +01002270
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002271===================================
2272INTER-CPU ACQUIRING BARRIER EFFECTS
2273===================================
David Howells108b42b2006-03-31 16:00:29 +01002274
2275On SMP systems locking primitives give a more substantial form of barrier: one
2276that does affect memory access ordering on other CPUs, within the context of
2277conflict on any particular lock.
2278
2279
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002280ACQUIRES VS MEMORY ACCESSES
2281---------------------------
David Howells108b42b2006-03-31 16:00:29 +01002282
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002283Consider the following: the system has a pair of spinlocks (M) and (Q), and
David Howells108b42b2006-03-31 16:00:29 +01002284three CPUs; then should the following sequence of events occur:
2285
2286 CPU 1 CPU 2
2287 =============================== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002288 WRITE_ONCE(*A, a); WRITE_ONCE(*E, e);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002289 ACQUIRE M ACQUIRE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002290 WRITE_ONCE(*B, b); WRITE_ONCE(*F, f);
2291 WRITE_ONCE(*C, c); WRITE_ONCE(*G, g);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002292 RELEASE M RELEASE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002293 WRITE_ONCE(*D, d); WRITE_ONCE(*H, h);
David Howells108b42b2006-03-31 16:00:29 +01002294
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002295Then there is no guarantee as to what order CPU 3 will see the accesses to *A
David Howells108b42b2006-03-31 16:00:29 +01002296through *H occur in, other than the constraints imposed by the separate locks
SeongJae Park0b6fa342016-04-12 08:52:53 -07002297on the separate CPUs. It might, for example, see:
David Howells108b42b2006-03-31 16:00:29 +01002298
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002299 *E, ACQUIRE M, ACQUIRE Q, *G, *C, *F, *A, *B, RELEASE Q, *D, *H, RELEASE M
David Howells108b42b2006-03-31 16:00:29 +01002300
2301But it won't see any of:
2302
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002303 *B, *C or *D preceding ACQUIRE M
2304 *A, *B or *C following RELEASE M
2305 *F, *G or *H preceding ACQUIRE Q
2306 *E, *F or *G following RELEASE Q
David Howells108b42b2006-03-31 16:00:29 +01002307
2308
David Howells108b42b2006-03-31 16:00:29 +01002309
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002310ACQUIRES VS I/O ACCESSES
2311------------------------
David Howells108b42b2006-03-31 16:00:29 +01002312
2313Under certain circumstances (especially involving NUMA), I/O accesses within
2314two spinlocked sections on two different CPUs may be seen as interleaved by the
2315PCI bridge, because the PCI bridge does not necessarily participate in the
2316cache-coherence protocol, and is therefore incapable of issuing the required
2317read memory barriers.
2318
2319For example:
2320
2321 CPU 1 CPU 2
2322 =============================== ===============================
2323 spin_lock(Q)
2324 writel(0, ADDR)
2325 writel(1, DATA);
2326 spin_unlock(Q);
2327 spin_lock(Q);
2328 writel(4, ADDR);
2329 writel(5, DATA);
2330 spin_unlock(Q);
2331
2332may be seen by the PCI bridge as follows:
2333
2334 STORE *ADDR = 0, STORE *ADDR = 4, STORE *DATA = 1, STORE *DATA = 5
2335
2336which would probably cause the hardware to malfunction.
2337
2338
2339What is necessary here is to intervene with an mmiowb() before dropping the
2340spinlock, for example:
2341
2342 CPU 1 CPU 2
2343 =============================== ===============================
2344 spin_lock(Q)
2345 writel(0, ADDR)
2346 writel(1, DATA);
2347 mmiowb();
2348 spin_unlock(Q);
2349 spin_lock(Q);
2350 writel(4, ADDR);
2351 writel(5, DATA);
2352 mmiowb();
2353 spin_unlock(Q);
2354
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002355this will ensure that the two stores issued on CPU 1 appear at the PCI bridge
2356before either of the stores issued on CPU 2.
David Howells108b42b2006-03-31 16:00:29 +01002357
2358
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002359Furthermore, following a store by a load from the same device obviates the need
2360for the mmiowb(), because the load forces the store to complete before the load
David Howells108b42b2006-03-31 16:00:29 +01002361is performed:
2362
2363 CPU 1 CPU 2
2364 =============================== ===============================
2365 spin_lock(Q)
2366 writel(0, ADDR)
2367 a = readl(DATA);
2368 spin_unlock(Q);
2369 spin_lock(Q);
2370 writel(4, ADDR);
2371 b = readl(DATA);
2372 spin_unlock(Q);
2373
2374
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002375See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002376
2377
2378=================================
2379WHERE ARE MEMORY BARRIERS NEEDED?
2380=================================
2381
2382Under normal operation, memory operation reordering is generally not going to
2383be a problem as a single-threaded linear piece of code will still appear to
David Howells50fa6102009-04-28 15:01:38 +01002384work correctly, even if it's in an SMP kernel. There are, however, four
David Howells108b42b2006-03-31 16:00:29 +01002385circumstances in which reordering definitely _could_ be a problem:
2386
2387 (*) Interprocessor interaction.
2388
2389 (*) Atomic operations.
2390
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002391 (*) Accessing devices.
David Howells108b42b2006-03-31 16:00:29 +01002392
2393 (*) Interrupts.
2394
2395
2396INTERPROCESSOR INTERACTION
2397--------------------------
2398
2399When there's a system with more than one processor, more than one CPU in the
2400system may be working on the same data set at the same time. This can cause
2401synchronisation problems, and the usual way of dealing with them is to use
2402locks. Locks, however, are quite expensive, and so it may be preferable to
2403operate without the use of a lock if at all possible. In such a case
2404operations that affect both CPUs may have to be carefully ordered to prevent
2405a malfunction.
2406
2407Consider, for example, the R/W semaphore slow path. Here a waiting process is
2408queued on the semaphore, by virtue of it having a piece of its stack linked to
2409the semaphore's list of waiting processes:
2410
2411 struct rw_semaphore {
2412 ...
2413 spinlock_t lock;
2414 struct list_head waiters;
2415 };
2416
2417 struct rwsem_waiter {
2418 struct list_head list;
2419 struct task_struct *task;
2420 };
2421
2422To wake up a particular waiter, the up_read() or up_write() functions have to:
2423
2424 (1) read the next pointer from this waiter's record to know as to where the
2425 next waiter record is;
2426
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002427 (2) read the pointer to the waiter's task structure;
David Howells108b42b2006-03-31 16:00:29 +01002428
2429 (3) clear the task pointer to tell the waiter it has been given the semaphore;
2430
2431 (4) call wake_up_process() on the task; and
2432
2433 (5) release the reference held on the waiter's task struct.
2434
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002435In other words, it has to perform this sequence of events:
David Howells108b42b2006-03-31 16:00:29 +01002436
2437 LOAD waiter->list.next;
2438 LOAD waiter->task;
2439 STORE waiter->task;
2440 CALL wakeup
2441 RELEASE task
2442
2443and if any of these steps occur out of order, then the whole thing may
2444malfunction.
2445
2446Once it has queued itself and dropped the semaphore lock, the waiter does not
2447get the lock again; it instead just waits for its task pointer to be cleared
2448before proceeding. Since the record is on the waiter's stack, this means that
2449if the task pointer is cleared _before_ the next pointer in the list is read,
2450another CPU might start processing the waiter and might clobber the waiter's
2451stack before the up*() function has a chance to read the next pointer.
2452
2453Consider then what might happen to the above sequence of events:
2454
2455 CPU 1 CPU 2
2456 =============================== ===============================
2457 down_xxx()
2458 Queue waiter
2459 Sleep
2460 up_yyy()
2461 LOAD waiter->task;
2462 STORE waiter->task;
2463 Woken up by other event
2464 <preempt>
2465 Resume processing
2466 down_xxx() returns
2467 call foo()
2468 foo() clobbers *waiter
2469 </preempt>
2470 LOAD waiter->list.next;
2471 --- OOPS ---
2472
2473This could be dealt with using the semaphore lock, but then the down_xxx()
2474function has to needlessly get the spinlock again after being woken up.
2475
2476The way to deal with this is to insert a general SMP memory barrier:
2477
2478 LOAD waiter->list.next;
2479 LOAD waiter->task;
2480 smp_mb();
2481 STORE waiter->task;
2482 CALL wakeup
2483 RELEASE task
2484
2485In this case, the barrier makes a guarantee that all memory accesses before the
2486barrier will appear to happen before all the memory accesses after the barrier
2487with respect to the other CPUs on the system. It does _not_ guarantee that all
2488the memory accesses before the barrier will be complete by the time the barrier
2489instruction itself is complete.
2490
2491On a UP system - where this wouldn't be a problem - the smp_mb() is just a
2492compiler barrier, thus making sure the compiler emits the instructions in the
David Howells6bc39272006-06-25 05:49:22 -07002493right order without actually intervening in the CPU. Since there's only one
2494CPU, that CPU's dependency ordering logic will take care of everything else.
David Howells108b42b2006-03-31 16:00:29 +01002495
2496
2497ATOMIC OPERATIONS
2498-----------------
2499
David Howellsdbc87002006-04-10 22:54:23 -07002500Whilst they are technically interprocessor interaction considerations, atomic
2501operations are noted specially as some of them imply full memory barriers and
2502some don't, but they're very heavily relied on as a group throughout the
2503kernel.
2504
Peter Zijlstra706eeb32017-06-12 14:50:27 +02002505See Documentation/atomic_t.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01002506
2507
2508ACCESSING DEVICES
2509-----------------
2510
2511Many devices can be memory mapped, and so appear to the CPU as if they're just
2512a set of memory locations. To control such a device, the driver usually has to
2513make the right memory accesses in exactly the right order.
2514
2515However, having a clever CPU or a clever compiler creates a potential problem
2516in that the carefully sequenced accesses in the driver code won't reach the
2517device in the requisite order if the CPU or the compiler thinks it is more
2518efficient to reorder, combine or merge accesses - something that would cause
2519the device to malfunction.
2520
2521Inside of the Linux kernel, I/O should be done through the appropriate accessor
2522routines - such as inb() or writel() - which know how to make such accesses
2523appropriately sequential. Whilst this, for the most part, renders the explicit
2524use of memory barriers unnecessary, there are a couple of situations where they
2525might be needed:
2526
2527 (1) On some systems, I/O stores are not strongly ordered across all CPUs, and
2528 so for _all_ general drivers locks should be used and mmiowb() must be
2529 issued prior to unlocking the critical section.
2530
2531 (2) If the accessor functions are used to refer to an I/O memory window with
2532 relaxed memory access properties, then _mandatory_ memory barriers are
2533 required to enforce ordering.
2534
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002535See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002536
2537
2538INTERRUPTS
2539----------
2540
2541A driver may be interrupted by its own interrupt service routine, and thus the
2542two parts of the driver may interfere with each other's attempts to control or
2543access the device.
2544
2545This may be alleviated - at least in part - by disabling local interrupts (a
2546form of locking), such that the critical operations are all contained within
2547the interrupt-disabled section in the driver. Whilst the driver's interrupt
2548routine is executing, the driver's core may not run on the same CPU, and its
2549interrupt is not permitted to happen again until the current interrupt has been
2550handled, thus the interrupt handler does not need to lock against that.
2551
2552However, consider a driver that was talking to an ethernet card that sports an
2553address register and a data register. If that driver's core talks to the card
2554under interrupt-disablement and then the driver's interrupt handler is invoked:
2555
2556 LOCAL IRQ DISABLE
2557 writew(ADDR, 3);
2558 writew(DATA, y);
2559 LOCAL IRQ ENABLE
2560 <interrupt>
2561 writew(ADDR, 4);
2562 q = readw(DATA);
2563 </interrupt>
2564
2565The store to the data register might happen after the second store to the
2566address register if ordering rules are sufficiently relaxed:
2567
2568 STORE *ADDR = 3, STORE *ADDR = 4, STORE *DATA = y, q = LOAD *DATA
2569
2570
2571If ordering rules are relaxed, it must be assumed that accesses done inside an
2572interrupt disabled section may leak outside of it and may interleave with
2573accesses performed in an interrupt - and vice versa - unless implicit or
2574explicit barriers are used.
2575
2576Normally this won't be a problem because the I/O accesses done inside such
2577sections will include synchronous load operations on strictly ordered I/O
SeongJae Park0b6fa342016-04-12 08:52:53 -07002578registers that form implicit I/O barriers. If this isn't sufficient then an
David Howells108b42b2006-03-31 16:00:29 +01002579mmiowb() may need to be used explicitly.
2580
2581
2582A similar situation may occur between an interrupt routine and two routines
SeongJae Park0b6fa342016-04-12 08:52:53 -07002583running on separate CPUs that communicate with each other. If such a case is
David Howells108b42b2006-03-31 16:00:29 +01002584likely, then interrupt-disabling locks should be used to guarantee ordering.
2585
2586
2587==========================
2588KERNEL I/O BARRIER EFFECTS
2589==========================
2590
2591When accessing I/O memory, drivers should use the appropriate accessor
2592functions:
2593
2594 (*) inX(), outX():
2595
2596 These are intended to talk to I/O space rather than memory space, but
SeongJae Park0b6fa342016-04-12 08:52:53 -07002597 that's primarily a CPU-specific concept. The i386 and x86_64 processors
2598 do indeed have special I/O space access cycles and instructions, but many
David Howells108b42b2006-03-31 16:00:29 +01002599 CPUs don't have such a concept.
2600
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002601 The PCI bus, amongst others, defines an I/O space concept which - on such
2602 CPUs as i386 and x86_64 - readily maps to the CPU's concept of I/O
David Howells6bc39272006-06-25 05:49:22 -07002603 space. However, it may also be mapped as a virtual I/O space in the CPU's
2604 memory map, particularly on those CPUs that don't support alternate I/O
2605 spaces.
David Howells108b42b2006-03-31 16:00:29 +01002606
2607 Accesses to this space may be fully synchronous (as on i386), but
2608 intermediary bridges (such as the PCI host bridge) may not fully honour
2609 that.
2610
2611 They are guaranteed to be fully ordered with respect to each other.
2612
2613 They are not guaranteed to be fully ordered with respect to other types of
2614 memory and I/O operation.
2615
2616 (*) readX(), writeX():
2617
2618 Whether these are guaranteed to be fully ordered and uncombined with
2619 respect to each other on the issuing CPU depends on the characteristics
SeongJae Park0b6fa342016-04-12 08:52:53 -07002620 defined for the memory window through which they're accessing. On later
David Howells108b42b2006-03-31 16:00:29 +01002621 i386 architecture machines, for example, this is controlled by way of the
2622 MTRR registers.
2623
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002624 Ordinarily, these will be guaranteed to be fully ordered and uncombined,
David Howells108b42b2006-03-31 16:00:29 +01002625 provided they're not accessing a prefetchable device.
2626
2627 However, intermediary hardware (such as a PCI bridge) may indulge in
2628 deferral if it so wishes; to flush a store, a load from the same location
2629 is preferred[*], but a load from the same device or from configuration
2630 space should suffice for PCI.
2631
2632 [*] NOTE! attempting to load from the same location as was written to may
Ingo Molnare0edc782013-11-22 11:24:53 +01002633 cause a malfunction - consider the 16550 Rx/Tx serial registers for
2634 example.
David Howells108b42b2006-03-31 16:00:29 +01002635
2636 Used with prefetchable I/O memory, an mmiowb() barrier may be required to
2637 force stores to be ordered.
2638
2639 Please refer to the PCI specification for more information on interactions
2640 between PCI transactions.
2641
Will Deacona8e0aea2013-09-04 12:30:08 +01002642 (*) readX_relaxed(), writeX_relaxed()
David Howells108b42b2006-03-31 16:00:29 +01002643
Will Deacona8e0aea2013-09-04 12:30:08 +01002644 These are similar to readX() and writeX(), but provide weaker memory
SeongJae Park0b6fa342016-04-12 08:52:53 -07002645 ordering guarantees. Specifically, they do not guarantee ordering with
Will Deacona8e0aea2013-09-04 12:30:08 +01002646 respect to normal memory accesses (e.g. DMA buffers) nor do they guarantee
SeongJae Park0b6fa342016-04-12 08:52:53 -07002647 ordering with respect to LOCK or UNLOCK operations. If the latter is
2648 required, an mmiowb() barrier can be used. Note that relaxed accesses to
Will Deacona8e0aea2013-09-04 12:30:08 +01002649 the same peripheral are guaranteed to be ordered with respect to each
2650 other.
David Howells108b42b2006-03-31 16:00:29 +01002651
2652 (*) ioreadX(), iowriteX()
2653
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002654 These will perform appropriately for the type of access they're actually
David Howells108b42b2006-03-31 16:00:29 +01002655 doing, be it inX()/outX() or readX()/writeX().
2656
2657
2658========================================
2659ASSUMED MINIMUM EXECUTION ORDERING MODEL
2660========================================
2661
2662It has to be assumed that the conceptual CPU is weakly-ordered but that it will
2663maintain the appearance of program causality with respect to itself. Some CPUs
2664(such as i386 or x86_64) are more constrained than others (such as powerpc or
2665frv), and so the most relaxed case (namely DEC Alpha) must be assumed outside
2666of arch-specific code.
2667
2668This means that it must be considered that the CPU will execute its instruction
2669stream in any order it feels like - or even in parallel - provided that if an
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002670instruction in the stream depends on an earlier instruction, then that
David Howells108b42b2006-03-31 16:00:29 +01002671earlier instruction must be sufficiently complete[*] before the later
2672instruction may proceed; in other words: provided that the appearance of
2673causality is maintained.
2674
2675 [*] Some instructions have more than one effect - such as changing the
2676 condition codes, changing registers or changing memory - and different
2677 instructions may depend on different effects.
2678
2679A CPU may also discard any instruction sequence that winds up having no
2680ultimate effect. For example, if two adjacent instructions both load an
2681immediate value into the same register, the first may be discarded.
2682
2683
2684Similarly, it has to be assumed that compiler might reorder the instruction
2685stream in any way it sees fit, again provided the appearance of causality is
2686maintained.
2687
2688
2689============================
2690THE EFFECTS OF THE CPU CACHE
2691============================
2692
2693The way cached memory operations are perceived across the system is affected to
2694a certain extent by the caches that lie between CPUs and memory, and by the
2695memory coherence system that maintains the consistency of state in the system.
2696
2697As far as the way a CPU interacts with another part of the system through the
2698caches goes, the memory system has to include the CPU's caches, and memory
2699barriers for the most part act at the interface between the CPU and its cache
2700(memory barriers logically act on the dotted line in the following diagram):
2701
2702 <--- CPU ---> : <----------- Memory ----------->
2703 :
2704 +--------+ +--------+ : +--------+ +-----------+
2705 | | | | : | | | | +--------+
Ingo Molnare0edc782013-11-22 11:24:53 +01002706 | CPU | | Memory | : | CPU | | | | |
2707 | Core |--->| Access |----->| Cache |<-->| | | |
David Howells108b42b2006-03-31 16:00:29 +01002708 | | | Queue | : | | | |--->| Memory |
Ingo Molnare0edc782013-11-22 11:24:53 +01002709 | | | | : | | | | | |
2710 +--------+ +--------+ : +--------+ | | | |
David Howells108b42b2006-03-31 16:00:29 +01002711 : | Cache | +--------+
2712 : | Coherency |
2713 : | Mechanism | +--------+
2714 +--------+ +--------+ : +--------+ | | | |
2715 | | | | : | | | | | |
2716 | CPU | | Memory | : | CPU | | |--->| Device |
Ingo Molnare0edc782013-11-22 11:24:53 +01002717 | Core |--->| Access |----->| Cache |<-->| | | |
2718 | | | Queue | : | | | | | |
David Howells108b42b2006-03-31 16:00:29 +01002719 | | | | : | | | | +--------+
2720 +--------+ +--------+ : +--------+ +-----------+
2721 :
2722 :
2723
2724Although any particular load or store may not actually appear outside of the
2725CPU that issued it since it may have been satisfied within the CPU's own cache,
2726it will still appear as if the full memory access had taken place as far as the
2727other CPUs are concerned since the cache coherency mechanisms will migrate the
2728cacheline over to the accessing CPU and propagate the effects upon conflict.
2729
2730The CPU core may execute instructions in any order it deems fit, provided the
2731expected program causality appears to be maintained. Some of the instructions
2732generate load and store operations which then go into the queue of memory
2733accesses to be performed. The core may place these in the queue in any order
2734it wishes, and continue execution until it is forced to wait for an instruction
2735to complete.
2736
2737What memory barriers are concerned with is controlling the order in which
2738accesses cross from the CPU side of things to the memory side of things, and
2739the order in which the effects are perceived to happen by the other observers
2740in the system.
2741
2742[!] Memory barriers are _not_ needed within a given CPU, as CPUs always see
2743their own loads and stores as if they had happened in program order.
2744
2745[!] MMIO or other device accesses may bypass the cache system. This depends on
2746the properties of the memory window through which devices are accessed and/or
2747the use of any special device communication instructions the CPU may have.
2748
2749
2750CACHE COHERENCY
2751---------------
2752
2753Life isn't quite as simple as it may appear above, however: for while the
2754caches are expected to be coherent, there's no guarantee that that coherency
2755will be ordered. This means that whilst changes made on one CPU will
2756eventually become visible on all CPUs, there's no guarantee that they will
2757become apparent in the same order on those other CPUs.
2758
2759
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002760Consider dealing with a system that has a pair of CPUs (1 & 2), each of which
2761has a pair of parallel data caches (CPU 1 has A/B, and CPU 2 has C/D):
David Howells108b42b2006-03-31 16:00:29 +01002762
2763 :
2764 : +--------+
2765 : +---------+ | |
2766 +--------+ : +--->| Cache A |<------->| |
2767 | | : | +---------+ | |
2768 | CPU 1 |<---+ | |
2769 | | : | +---------+ | |
2770 +--------+ : +--->| Cache B |<------->| |
2771 : +---------+ | |
2772 : | Memory |
2773 : +---------+ | System |
2774 +--------+ : +--->| Cache C |<------->| |
2775 | | : | +---------+ | |
2776 | CPU 2 |<---+ | |
2777 | | : | +---------+ | |
2778 +--------+ : +--->| Cache D |<------->| |
2779 : +---------+ | |
2780 : +--------+
2781 :
2782
2783Imagine the system has the following properties:
2784
2785 (*) an odd-numbered cache line may be in cache A, cache C or it may still be
2786 resident in memory;
2787
2788 (*) an even-numbered cache line may be in cache B, cache D or it may still be
2789 resident in memory;
2790
2791 (*) whilst the CPU core is interrogating one cache, the other cache may be
2792 making use of the bus to access the rest of the system - perhaps to
2793 displace a dirty cacheline or to do a speculative load;
2794
2795 (*) each cache has a queue of operations that need to be applied to that cache
2796 to maintain coherency with the rest of the system;
2797
2798 (*) the coherency queue is not flushed by normal loads to lines already
2799 present in the cache, even though the contents of the queue may
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002800 potentially affect those loads.
David Howells108b42b2006-03-31 16:00:29 +01002801
2802Imagine, then, that two writes are made on the first CPU, with a write barrier
2803between them to guarantee that they will appear to reach that CPU's caches in
2804the requisite order:
2805
2806 CPU 1 CPU 2 COMMENT
2807 =============== =============== =======================================
2808 u == 0, v == 1 and p == &u, q == &u
2809 v = 2;
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002810 smp_wmb(); Make sure change to v is visible before
David Howells108b42b2006-03-31 16:00:29 +01002811 change to p
2812 <A:modify v=2> v is now in cache A exclusively
2813 p = &v;
2814 <B:modify p=&v> p is now in cache B exclusively
2815
2816The write memory barrier forces the other CPUs in the system to perceive that
2817the local CPU's caches have apparently been updated in the correct order. But
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002818now imagine that the second CPU wants to read those values:
David Howells108b42b2006-03-31 16:00:29 +01002819
2820 CPU 1 CPU 2 COMMENT
2821 =============== =============== =======================================
2822 ...
2823 q = p;
2824 x = *q;
2825
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002826The above pair of reads may then fail to happen in the expected order, as the
David Howells108b42b2006-03-31 16:00:29 +01002827cacheline holding p may get updated in one of the second CPU's caches whilst
2828the update to the cacheline holding v is delayed in the other of the second
2829CPU's caches by some other cache event:
2830
2831 CPU 1 CPU 2 COMMENT
2832 =============== =============== =======================================
2833 u == 0, v == 1 and p == &u, q == &u
2834 v = 2;
2835 smp_wmb();
2836 <A:modify v=2> <C:busy>
2837 <C:queue v=2>
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002838 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002839 <D:request p>
2840 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002841 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002842 x = *q;
2843 <C:read *q> Reads from v before v updated in cache
2844 <C:unbusy>
2845 <C:commit v=2>
2846
2847Basically, whilst both cachelines will be updated on CPU 2 eventually, there's
2848no guarantee that, without intervention, the order of update will be the same
2849as that committed on CPU 1.
2850
2851
2852To intervene, we need to interpolate a data dependency barrier or a read
Paul E. McKenneyf28f0862018-03-07 09:27:37 -08002853barrier between the loads (which as of v4.15 is supplied unconditionally
2854by the READ_ONCE() macro). This will force the cache to commit its
2855coherency queue before processing any further requests:
David Howells108b42b2006-03-31 16:00:29 +01002856
2857 CPU 1 CPU 2 COMMENT
2858 =============== =============== =======================================
2859 u == 0, v == 1 and p == &u, q == &u
2860 v = 2;
2861 smp_wmb();
2862 <A:modify v=2> <C:busy>
2863 <C:queue v=2>
Paolo 'Blaisorblade' Giarrusso3fda9822006-10-19 23:28:19 -07002864 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002865 <D:request p>
2866 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002867 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002868 smp_read_barrier_depends()
2869 <C:unbusy>
2870 <C:commit v=2>
2871 x = *q;
2872 <C:read *q> Reads from v after v updated in cache
2873
2874
2875This sort of problem can be encountered on DEC Alpha processors as they have a
2876split cache that improves performance by making better use of the data bus.
2877Whilst most CPUs do imply a data dependency barrier on the read when a memory
2878access depends on a read, not all do, so it may not be relied on.
2879
2880Other CPUs may also have split caches, but must coordinate between the various
Matt LaPlante3f6dee92006-10-03 22:45:33 +02002881cachelets for normal memory accesses. The semantics of the Alpha removes the
Paul E. McKenney9ad3c142017-11-27 09:20:40 -08002882need for hardware coordination in the absence of memory barriers, which
2883permitted Alpha to sport higher CPU clock rates back in the day. However,
Paul E. McKenneyf28f0862018-03-07 09:27:37 -08002884please note that (again, as of v4.15) smp_read_barrier_depends() should not
2885be used except in Alpha arch-specific code and within the READ_ONCE() macro.
David Howells108b42b2006-03-31 16:00:29 +01002886
2887
2888CACHE COHERENCY VS DMA
2889----------------------
2890
2891Not all systems maintain cache coherency with respect to devices doing DMA. In
2892such cases, a device attempting DMA may obtain stale data from RAM because
2893dirty cache lines may be resident in the caches of various CPUs, and may not
2894have been written back to RAM yet. To deal with this, the appropriate part of
2895the kernel must flush the overlapping bits of cache on each CPU (and maybe
2896invalidate them as well).
2897
2898In addition, the data DMA'd to RAM by a device may be overwritten by dirty
2899cache lines being written back to RAM from a CPU's cache after the device has
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002900installed its own data, or cache lines present in the CPU's cache may simply
2901obscure the fact that RAM has been updated, until at such time as the cacheline
2902is discarded from the CPU's cache and reloaded. To deal with this, the
2903appropriate part of the kernel must invalidate the overlapping bits of the
David Howells108b42b2006-03-31 16:00:29 +01002904cache on each CPU.
2905
2906See Documentation/cachetlb.txt for more information on cache management.
2907
2908
2909CACHE COHERENCY VS MMIO
2910-----------------------
2911
2912Memory mapped I/O usually takes place through memory locations that are part of
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002913a window in the CPU's memory space that has different properties assigned than
David Howells108b42b2006-03-31 16:00:29 +01002914the usual RAM directed window.
2915
2916Amongst these properties is usually the fact that such accesses bypass the
2917caching entirely and go directly to the device buses. This means MMIO accesses
2918may, in effect, overtake accesses to cached memory that were emitted earlier.
2919A memory barrier isn't sufficient in such a case, but rather the cache must be
2920flushed between the cached memory write and the MMIO access if the two are in
2921any way dependent.
2922
2923
2924=========================
2925THE THINGS CPUS GET UP TO
2926=========================
2927
2928A programmer might take it for granted that the CPU will perform memory
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002929operations in exactly the order specified, so that if the CPU is, for example,
David Howells108b42b2006-03-31 16:00:29 +01002930given the following piece of code to execute:
2931
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002932 a = READ_ONCE(*A);
2933 WRITE_ONCE(*B, b);
2934 c = READ_ONCE(*C);
2935 d = READ_ONCE(*D);
2936 WRITE_ONCE(*E, e);
David Howells108b42b2006-03-31 16:00:29 +01002937
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002938they would then expect that the CPU will complete the memory operation for each
David Howells108b42b2006-03-31 16:00:29 +01002939instruction before moving on to the next one, leading to a definite sequence of
2940operations as seen by external observers in the system:
2941
2942 LOAD *A, STORE *B, LOAD *C, LOAD *D, STORE *E.
2943
2944
2945Reality is, of course, much messier. With many CPUs and compilers, the above
2946assumption doesn't hold because:
2947
2948 (*) loads are more likely to need to be completed immediately to permit
2949 execution progress, whereas stores can often be deferred without a
2950 problem;
2951
2952 (*) loads may be done speculatively, and the result discarded should it prove
2953 to have been unnecessary;
2954
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002955 (*) loads may be done speculatively, leading to the result having been fetched
2956 at the wrong time in the expected sequence of events;
David Howells108b42b2006-03-31 16:00:29 +01002957
2958 (*) the order of the memory accesses may be rearranged to promote better use
2959 of the CPU buses and caches;
2960
2961 (*) loads and stores may be combined to improve performance when talking to
2962 memory or I/O hardware that can do batched accesses of adjacent locations,
2963 thus cutting down on transaction setup costs (memory and PCI devices may
2964 both be able to do this); and
2965
2966 (*) the CPU's data cache may affect the ordering, and whilst cache-coherency
2967 mechanisms may alleviate this - once the store has actually hit the cache
2968 - there's no guarantee that the coherency management will be propagated in
2969 order to other CPUs.
2970
2971So what another CPU, say, might actually observe from the above piece of code
2972is:
2973
2974 LOAD *A, ..., LOAD {*C,*D}, STORE *E, STORE *B
2975
2976 (Where "LOAD {*C,*D}" is a combined load)
2977
2978
2979However, it is guaranteed that a CPU will be self-consistent: it will see its
2980_own_ accesses appear to be correctly ordered, without the need for a memory
2981barrier. For instance with the following code:
2982
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002983 U = READ_ONCE(*A);
2984 WRITE_ONCE(*A, V);
2985 WRITE_ONCE(*A, W);
2986 X = READ_ONCE(*A);
2987 WRITE_ONCE(*A, Y);
2988 Z = READ_ONCE(*A);
David Howells108b42b2006-03-31 16:00:29 +01002989
2990and assuming no intervention by an external influence, it can be assumed that
2991the final result will appear to be:
2992
2993 U == the original value of *A
2994 X == W
2995 Z == Y
2996 *A == Y
2997
2998The code above may cause the CPU to generate the full sequence of memory
2999accesses:
3000
3001 U=LOAD *A, STORE *A=V, STORE *A=W, X=LOAD *A, STORE *A=Y, Z=LOAD *A
3002
3003in that order, but, without intervention, the sequence may have almost any
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003004combination of elements combined or discarded, provided the program's view
3005of the world remains consistent. Note that READ_ONCE() and WRITE_ONCE()
3006are -not- optional in the above example, as there are architectures
3007where a given CPU might reorder successive loads to the same location.
3008On such architectures, READ_ONCE() and WRITE_ONCE() do whatever is
3009necessary to prevent this, for example, on Itanium the volatile casts
3010used by READ_ONCE() and WRITE_ONCE() cause GCC to emit the special ld.acq
3011and st.rel instructions (respectively) that prevent such reordering.
David Howells108b42b2006-03-31 16:00:29 +01003012
3013The compiler may also combine, discard or defer elements of the sequence before
3014the CPU even sees them.
3015
3016For instance:
3017
3018 *A = V;
3019 *A = W;
3020
3021may be reduced to:
3022
3023 *A = W;
3024
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003025since, without either a write barrier or an WRITE_ONCE(), it can be
Paul E. McKenney2ecf8102013-12-11 13:59:04 -08003026assumed that the effect of the storage of V to *A is lost. Similarly:
David Howells108b42b2006-03-31 16:00:29 +01003027
3028 *A = Y;
3029 Z = *A;
3030
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003031may, without a memory barrier or an READ_ONCE() and WRITE_ONCE(), be
3032reduced to:
David Howells108b42b2006-03-31 16:00:29 +01003033
3034 *A = Y;
3035 Z = Y;
3036
3037and the LOAD operation never appear outside of the CPU.
3038
3039
3040AND THEN THERE'S THE ALPHA
3041--------------------------
3042
3043The DEC Alpha CPU is one of the most relaxed CPUs there is. Not only that,
3044some versions of the Alpha CPU have a split data cache, permitting them to have
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003045two semantically-related cache lines updated at separate times. This is where
David Howells108b42b2006-03-31 16:00:29 +01003046the data dependency barrier really becomes necessary as this synchronises both
3047caches with the memory coherence system, thus making it seem like pointer
3048changes vs new data occur in the right order.
3049
Paul E. McKenneyf28f0862018-03-07 09:27:37 -08003050The Alpha defines the Linux kernel's memory model, although as of v4.15
3051the Linux kernel's addition of smp_read_barrier_depends() to READ_ONCE()
3052greatly reduced Alpha's impact on the memory model.
David Howells108b42b2006-03-31 16:00:29 +01003053
3054See the subsection on "Cache Coherency" above.
3055
SeongJae Park0b6fa342016-04-12 08:52:53 -07003056
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003057VIRTUAL MACHINE GUESTS
SeongJae Park3dbf0912016-04-12 08:52:52 -07003058----------------------
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003059
3060Guests running within virtual machines might be affected by SMP effects even if
3061the guest itself is compiled without SMP support. This is an artifact of
3062interfacing with an SMP host while running an UP kernel. Using mandatory
3063barriers for this use-case would be possible but is often suboptimal.
3064
3065To handle this case optimally, low-level virt_mb() etc macros are available.
3066These have the same effect as smp_mb() etc when SMP is enabled, but generate
SeongJae Park0b6fa342016-04-12 08:52:53 -07003067identical code for SMP and non-SMP systems. For example, virtual machine guests
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003068should use virt_mb() rather than smp_mb() when synchronizing against a
3069(possibly SMP) host.
3070
3071These are equivalent to smp_mb() etc counterparts in all other respects,
3072in particular, they do not control MMIO effects: to control
3073MMIO effects, use mandatory barriers.
David Howells108b42b2006-03-31 16:00:29 +01003074
SeongJae Park0b6fa342016-04-12 08:52:53 -07003075
David Howells90fddab2010-03-24 09:43:00 +00003076============
3077EXAMPLE USES
3078============
3079
3080CIRCULAR BUFFERS
3081----------------
3082
3083Memory barriers can be used to implement circular buffering without the need
3084of a lock to serialise the producer with the consumer. See:
3085
3086 Documentation/circular-buffers.txt
3087
3088for details.
3089
3090
David Howells108b42b2006-03-31 16:00:29 +01003091==========
3092REFERENCES
3093==========
3094
3095Alpha AXP Architecture Reference Manual, Second Edition (Sites & Witek,
3096Digital Press)
3097 Chapter 5.2: Physical Address Space Characteristics
3098 Chapter 5.4: Caches and Write Buffers
3099 Chapter 5.5: Data Sharing
3100 Chapter 5.6: Read/Write Ordering
3101
3102AMD64 Architecture Programmer's Manual Volume 2: System Programming
3103 Chapter 7.1: Memory-Access Ordering
3104 Chapter 7.4: Buffering and Combining Memory Writes
3105
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07003106ARM Architecture Reference Manual (ARMv8, for ARMv8-A architecture profile)
3107 Chapter B2: The AArch64 Application Level Memory Model
3108
David Howells108b42b2006-03-31 16:00:29 +01003109IA-32 Intel Architecture Software Developer's Manual, Volume 3:
3110System Programming Guide
3111 Chapter 7.1: Locked Atomic Operations
3112 Chapter 7.2: Memory Ordering
3113 Chapter 7.4: Serializing Instructions
3114
3115The SPARC Architecture Manual, Version 9
3116 Chapter 8: Memory Models
3117 Appendix D: Formal Specification of the Memory Models
3118 Appendix J: Programming with the Memory Models
3119
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07003120Storage in the PowerPC (Stone and Fitzgerald)
3121
David Howells108b42b2006-03-31 16:00:29 +01003122UltraSPARC Programmer Reference Manual
3123 Chapter 5: Memory Accesses and Cacheability
3124 Chapter 15: Sparc-V9 Memory Models
3125
3126UltraSPARC III Cu User's Manual
3127 Chapter 9: Memory Models
3128
3129UltraSPARC IIIi Processor User's Manual
3130 Chapter 8: Memory Models
3131
3132UltraSPARC Architecture 2005
3133 Chapter 9: Memory
3134 Appendix D: Formal Specifications of the Memory Models
3135
3136UltraSPARC T1 Supplement to the UltraSPARC Architecture 2005
3137 Chapter 8: Memory Models
3138 Appendix F: Caches and Cache Coherency
3139
3140Solaris Internals, Core Kernel Architecture, p63-68:
3141 Chapter 3.3: Hardware Considerations for Locks and
3142 Synchronization
3143
3144Unix Systems for Modern Architectures, Symmetric Multiprocessing and Caching
3145for Kernel Programmers:
3146 Chapter 13: Other Memory Models
3147
3148Intel Itanium Architecture Software Developer's Manual: Volume 1:
3149 Section 2.6: Speculation
3150 Section 4.4: Memory Access