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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
17in case of any doubt (and there are many) please ask.
18
19To repeat, this document is not a specification of what Linux expects from
20hardware.
21
David Howells8d4840e2016-04-26 10:22:06 -070022The purpose of this document is twofold:
23
24 (1) to specify the minimum functionality that one can rely on for any
25 particular barrier, and
26
27 (2) to provide a guide as to how to use the barriers that are available.
28
29Note that an architecture can provide more than the minimum requirement
Stan Drozd35bdc722017-04-20 11:03:36 +020030for any particular barrier, but if the architecture provides less than
David Howells8d4840e2016-04-26 10:22:06 -070031that, that architecture is incorrect.
32
33Note also that it is possible that a barrier may be a no-op for an
34architecture because the way that arch works renders an explicit barrier
35unnecessary in that case.
36
37
Peter Zijlstrae7720af2016-04-26 10:22:05 -070038========
39CONTENTS
40========
David Howells108b42b2006-03-31 16:00:29 +010041
42 (*) Abstract memory access model.
43
44 - Device operations.
45 - Guarantees.
46
47 (*) What are memory barriers?
48
49 - Varieties of memory barrier.
50 - What may not be assumed about memory barriers?
51 - Data dependency barriers.
52 - Control dependencies.
53 - SMP barrier pairing.
54 - Examples of memory barrier sequences.
David Howells670bd952006-06-10 09:54:12 -070055 - Read memory barriers vs load speculation.
Paul E. McKenney241e6662011-02-10 16:54:50 -080056 - Transitivity
David Howells108b42b2006-03-31 16:00:29 +010057
58 (*) Explicit kernel barriers.
59
60 - Compiler barrier.
Jarek Poplawski81fc6322007-05-23 13:58:20 -070061 - CPU memory barriers.
David Howells108b42b2006-03-31 16:00:29 +010062 - MMIO write barrier.
63
64 (*) Implicit kernel memory barriers.
65
SeongJae Park166bda72016-04-12 08:52:50 -070066 - Lock acquisition functions.
David Howells108b42b2006-03-31 16:00:29 +010067 - Interrupt disabling functions.
David Howells50fa6102009-04-28 15:01:38 +010068 - Sleep and wake-up functions.
David Howells108b42b2006-03-31 16:00:29 +010069 - Miscellaneous functions.
70
SeongJae Park166bda72016-04-12 08:52:50 -070071 (*) Inter-CPU acquiring barrier effects.
David Howells108b42b2006-03-31 16:00:29 +010072
SeongJae Park166bda72016-04-12 08:52:50 -070073 - Acquires vs memory accesses.
74 - Acquires vs I/O accesses.
David Howells108b42b2006-03-31 16:00:29 +010075
76 (*) Where are memory barriers needed?
77
78 - Interprocessor interaction.
79 - Atomic operations.
80 - Accessing devices.
81 - Interrupts.
82
83 (*) Kernel I/O barrier effects.
84
85 (*) Assumed minimum execution ordering model.
86
87 (*) The effects of the cpu cache.
88
89 - Cache coherency.
90 - Cache coherency vs DMA.
91 - Cache coherency vs MMIO.
92
93 (*) The things CPUs get up to.
94
95 - And then there's the Alpha.
SeongJae Park01e1cd62016-04-12 08:52:51 -070096 - Virtual Machine Guests.
David Howells108b42b2006-03-31 16:00:29 +010097
David Howells90fddab2010-03-24 09:43:00 +000098 (*) Example uses.
99
100 - Circular buffers.
101
David Howells108b42b2006-03-31 16:00:29 +0100102 (*) References.
103
104
105============================
106ABSTRACT MEMORY ACCESS MODEL
107============================
108
109Consider the following abstract model of the system:
110
111 : :
112 : :
113 : :
114 +-------+ : +--------+ : +-------+
115 | | : | | : | |
116 | | : | | : | |
117 | CPU 1 |<----->| Memory |<----->| CPU 2 |
118 | | : | | : | |
119 | | : | | : | |
120 +-------+ : +--------+ : +-------+
121 ^ : ^ : ^
122 | : | : |
123 | : | : |
124 | : v : |
125 | : +--------+ : |
126 | : | | : |
127 | : | | : |
128 +---------->| Device |<----------+
129 : | | :
130 : | | :
131 : +--------+ :
132 : :
133
134Each CPU executes a program that generates memory access operations. In the
135abstract CPU, memory operation ordering is very relaxed, and a CPU may actually
136perform the memory operations in any order it likes, provided program causality
137appears to be maintained. Similarly, the compiler may also arrange the
138instructions it emits in any order it likes, provided it doesn't affect the
139apparent operation of the program.
140
141So in the above diagram, the effects of the memory operations performed by a
142CPU are perceived by the rest of the system as the operations cross the
143interface between the CPU and rest of the system (the dotted lines).
144
145
146For example, consider the following sequence of events:
147
148 CPU 1 CPU 2
149 =============== ===============
150 { A == 1; B == 2 }
Alexey Dobriyan615cc2c2014-06-06 14:36:41 -0700151 A = 3; x = B;
152 B = 4; y = A;
David Howells108b42b2006-03-31 16:00:29 +0100153
154The set of accesses as seen by the memory system in the middle can be arranged
155in 24 different combinations:
156
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400157 STORE A=3, STORE B=4, y=LOAD A->3, x=LOAD B->4
158 STORE A=3, STORE B=4, x=LOAD B->4, y=LOAD A->3
159 STORE A=3, y=LOAD A->3, STORE B=4, x=LOAD B->4
160 STORE A=3, y=LOAD A->3, x=LOAD B->2, STORE B=4
161 STORE A=3, x=LOAD B->2, STORE B=4, y=LOAD A->3
162 STORE A=3, x=LOAD B->2, y=LOAD A->3, STORE B=4
163 STORE B=4, STORE A=3, y=LOAD A->3, x=LOAD B->4
David Howells108b42b2006-03-31 16:00:29 +0100164 STORE B=4, ...
165 ...
166
167and can thus result in four different combinations of values:
168
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400169 x == 2, y == 1
170 x == 2, y == 3
171 x == 4, y == 1
172 x == 4, y == 3
David Howells108b42b2006-03-31 16:00:29 +0100173
174
175Furthermore, the stores committed by a CPU to the memory system may not be
176perceived by the loads made by another CPU in the same order as the stores were
177committed.
178
179
180As a further example, consider this sequence of events:
181
182 CPU 1 CPU 2
183 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700184 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100185 B = 4; Q = P;
186 P = &B D = *Q;
187
188There is an obvious data dependency here, as the value loaded into D depends on
189the address retrieved from P by CPU 2. At the end of the sequence, any of the
190following results are possible:
191
192 (Q == &A) and (D == 1)
193 (Q == &B) and (D == 2)
194 (Q == &B) and (D == 4)
195
196Note that CPU 2 will never try and load C into D because the CPU will load P
197into Q before issuing the load of *Q.
198
199
200DEVICE OPERATIONS
201-----------------
202
203Some devices present their control interfaces as collections of memory
204locations, but the order in which the control registers are accessed is very
205important. For instance, imagine an ethernet card with a set of internal
206registers that are accessed through an address port register (A) and a data
207port register (D). To read internal register 5, the following code might then
208be used:
209
210 *A = 5;
211 x = *D;
212
213but this might show up as either of the following two sequences:
214
215 STORE *A = 5, x = LOAD *D
216 x = LOAD *D, STORE *A = 5
217
218the second of which will almost certainly result in a malfunction, since it set
219the address _after_ attempting to read the register.
220
221
222GUARANTEES
223----------
224
225There are some minimal guarantees that may be expected of a CPU:
226
227 (*) On any given CPU, dependent memory accesses will be issued in order, with
228 respect to itself. This means that for:
229
Chris Metcalff84cfbb2015-11-23 17:04:17 -0500230 Q = READ_ONCE(P); smp_read_barrier_depends(); D = READ_ONCE(*Q);
David Howells108b42b2006-03-31 16:00:29 +0100231
232 the CPU will issue the following memory operations:
233
234 Q = LOAD P, D = LOAD *Q
235
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800236 and always in that order. On most systems, smp_read_barrier_depends()
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700237 does nothing, but it is required for DEC Alpha. The READ_ONCE()
Chris Metcalff84cfbb2015-11-23 17:04:17 -0500238 is required to prevent compiler mischief. Please note that you
239 should normally use something like rcu_dereference() instead of
240 open-coding smp_read_barrier_depends().
David Howells108b42b2006-03-31 16:00:29 +0100241
242 (*) Overlapping loads and stores within a particular CPU will appear to be
243 ordered within that CPU. This means that for:
244
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700245 a = READ_ONCE(*X); WRITE_ONCE(*X, b);
David Howells108b42b2006-03-31 16:00:29 +0100246
247 the CPU will only issue the following sequence of memory operations:
248
249 a = LOAD *X, STORE *X = b
250
251 And for:
252
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700253 WRITE_ONCE(*X, c); d = READ_ONCE(*X);
David Howells108b42b2006-03-31 16:00:29 +0100254
255 the CPU will only issue:
256
257 STORE *X = c, d = LOAD *X
258
Matt LaPlantefa00e7e2006-11-30 04:55:36 +0100259 (Loads and stores overlap if they are targeted at overlapping pieces of
David Howells108b42b2006-03-31 16:00:29 +0100260 memory).
261
262And there are a number of things that _must_ or _must_not_ be assumed:
263
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700264 (*) It _must_not_ be assumed that the compiler will do what you want
265 with memory references that are not protected by READ_ONCE() and
266 WRITE_ONCE(). Without them, the compiler is within its rights to
267 do all sorts of "creative" transformations, which are covered in
Paul E. McKenney895f5542016-01-06 14:23:03 -0800268 the COMPILER BARRIER section.
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800269
David Howells108b42b2006-03-31 16:00:29 +0100270 (*) It _must_not_ be assumed that independent loads and stores will be issued
271 in the order given. This means that for:
272
273 X = *A; Y = *B; *D = Z;
274
275 we may get any of the following sequences:
276
277 X = LOAD *A, Y = LOAD *B, STORE *D = Z
278 X = LOAD *A, STORE *D = Z, Y = LOAD *B
279 Y = LOAD *B, X = LOAD *A, STORE *D = Z
280 Y = LOAD *B, STORE *D = Z, X = LOAD *A
281 STORE *D = Z, X = LOAD *A, Y = LOAD *B
282 STORE *D = Z, Y = LOAD *B, X = LOAD *A
283
284 (*) It _must_ be assumed that overlapping memory accesses may be merged or
285 discarded. This means that for:
286
287 X = *A; Y = *(A + 4);
288
289 we may get any one of the following sequences:
290
291 X = LOAD *A; Y = LOAD *(A + 4);
292 Y = LOAD *(A + 4); X = LOAD *A;
293 {X, Y} = LOAD {*A, *(A + 4) };
294
295 And for:
296
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700297 *A = X; *(A + 4) = Y;
David Howells108b42b2006-03-31 16:00:29 +0100298
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700299 we may get any of:
David Howells108b42b2006-03-31 16:00:29 +0100300
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700301 STORE *A = X; STORE *(A + 4) = Y;
302 STORE *(A + 4) = Y; STORE *A = X;
303 STORE {*A, *(A + 4) } = {X, Y};
David Howells108b42b2006-03-31 16:00:29 +0100304
Paul E. McKenney432fbf32014-09-04 17:12:49 -0700305And there are anti-guarantees:
306
307 (*) These guarantees do not apply to bitfields, because compilers often
308 generate code to modify these using non-atomic read-modify-write
309 sequences. Do not attempt to use bitfields to synchronize parallel
310 algorithms.
311
312 (*) Even in cases where bitfields are protected by locks, all fields
313 in a given bitfield must be protected by one lock. If two fields
314 in a given bitfield are protected by different locks, the compiler's
315 non-atomic read-modify-write sequences can cause an update to one
316 field to corrupt the value of an adjacent field.
317
318 (*) These guarantees apply only to properly aligned and sized scalar
319 variables. "Properly sized" currently means variables that are
320 the same size as "char", "short", "int" and "long". "Properly
321 aligned" means the natural alignment, thus no constraints for
322 "char", two-byte alignment for "short", four-byte alignment for
323 "int", and either four-byte or eight-byte alignment for "long",
324 on 32-bit and 64-bit systems, respectively. Note that these
325 guarantees were introduced into the C11 standard, so beware when
326 using older pre-C11 compilers (for example, gcc 4.6). The portion
327 of the standard containing this guarantee is Section 3.14, which
328 defines "memory location" as follows:
329
330 memory location
331 either an object of scalar type, or a maximal sequence
332 of adjacent bit-fields all having nonzero width
333
334 NOTE 1: Two threads of execution can update and access
335 separate memory locations without interfering with
336 each other.
337
338 NOTE 2: A bit-field and an adjacent non-bit-field member
339 are in separate memory locations. The same applies
340 to two bit-fields, if one is declared inside a nested
341 structure declaration and the other is not, or if the two
342 are separated by a zero-length bit-field declaration,
343 or if they are separated by a non-bit-field member
344 declaration. It is not safe to concurrently update two
345 bit-fields in the same structure if all members declared
346 between them are also bit-fields, no matter what the
347 sizes of those intervening bit-fields happen to be.
348
David Howells108b42b2006-03-31 16:00:29 +0100349
350=========================
351WHAT ARE MEMORY BARRIERS?
352=========================
353
354As can be seen above, independent memory operations are effectively performed
355in random order, but this can be a problem for CPU-CPU interaction and for I/O.
356What is required is some way of intervening to instruct the compiler and the
357CPU to restrict the order.
358
359Memory barriers are such interventions. They impose a perceived partial
David Howells2b948952006-06-25 05:48:49 -0700360ordering over the memory operations on either side of the barrier.
361
362Such enforcement is important because the CPUs and other devices in a system
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700363can use a variety of tricks to improve performance, including reordering,
David Howells2b948952006-06-25 05:48:49 -0700364deferral and combination of memory operations; speculative loads; speculative
365branch prediction and various types of caching. Memory barriers are used to
366override or suppress these tricks, allowing the code to sanely control the
367interaction of multiple CPUs and/or devices.
David Howells108b42b2006-03-31 16:00:29 +0100368
369
370VARIETIES OF MEMORY BARRIER
371---------------------------
372
373Memory barriers come in four basic varieties:
374
375 (1) Write (or store) memory barriers.
376
377 A write memory barrier gives a guarantee that all the STORE operations
378 specified before the barrier will appear to happen before all the STORE
379 operations specified after the barrier with respect to the other
380 components of the system.
381
382 A write barrier is a partial ordering on stores only; it is not required
383 to have any effect on loads.
384
David Howells6bc39272006-06-25 05:49:22 -0700385 A CPU can be viewed as committing a sequence of store operations to the
David Howells108b42b2006-03-31 16:00:29 +0100386 memory system as time progresses. All stores before a write barrier will
387 occur in the sequence _before_ all the stores after the write barrier.
388
389 [!] Note that write barriers should normally be paired with read or data
390 dependency barriers; see the "SMP barrier pairing" subsection.
391
392
393 (2) Data dependency barriers.
394
395 A data dependency barrier is a weaker form of read barrier. In the case
396 where two loads are performed such that the second depends on the result
397 of the first (eg: the first load retrieves the address to which the second
398 load will be directed), a data dependency barrier would be required to
399 make sure that the target of the second load is updated before the address
400 obtained by the first load is accessed.
401
402 A data dependency barrier is a partial ordering on interdependent loads
403 only; it is not required to have any effect on stores, independent loads
404 or overlapping loads.
405
406 As mentioned in (1), the other CPUs in the system can be viewed as
407 committing sequences of stores to the memory system that the CPU being
408 considered can then perceive. A data dependency barrier issued by the CPU
409 under consideration guarantees that for any load preceding it, if that
410 load touches one of a sequence of stores from another CPU, then by the
411 time the barrier completes, the effects of all the stores prior to that
412 touched by the load will be perceptible to any loads issued after the data
413 dependency barrier.
414
415 See the "Examples of memory barrier sequences" subsection for diagrams
416 showing the ordering constraints.
417
418 [!] Note that the first load really has to have a _data_ dependency and
419 not a control dependency. If the address for the second load is dependent
420 on the first load, but the dependency is through a conditional rather than
421 actually loading the address itself, then it's a _control_ dependency and
422 a full read barrier or better is required. See the "Control dependencies"
423 subsection for more information.
424
425 [!] Note that data dependency barriers should normally be paired with
426 write barriers; see the "SMP barrier pairing" subsection.
427
428
429 (3) Read (or load) memory barriers.
430
431 A read barrier is a data dependency barrier plus a guarantee that all the
432 LOAD operations specified before the barrier will appear to happen before
433 all the LOAD operations specified after the barrier with respect to the
434 other components of the system.
435
436 A read barrier is a partial ordering on loads only; it is not required to
437 have any effect on stores.
438
439 Read memory barriers imply data dependency barriers, and so can substitute
440 for them.
441
442 [!] Note that read barriers should normally be paired with write barriers;
443 see the "SMP barrier pairing" subsection.
444
445
446 (4) General memory barriers.
447
David Howells670bd952006-06-10 09:54:12 -0700448 A general memory barrier gives a guarantee that all the LOAD and STORE
449 operations specified before the barrier will appear to happen before all
450 the LOAD and STORE operations specified after the barrier with respect to
451 the other components of the system.
452
453 A general memory barrier is a partial ordering over both loads and stores.
David Howells108b42b2006-03-31 16:00:29 +0100454
455 General memory barriers imply both read and write memory barriers, and so
456 can substitute for either.
457
458
459And a couple of implicit varieties:
460
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100461 (5) ACQUIRE operations.
David Howells108b42b2006-03-31 16:00:29 +0100462
463 This acts as a one-way permeable barrier. It guarantees that all memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100464 operations after the ACQUIRE operation will appear to happen after the
465 ACQUIRE operation with respect to the other components of the system.
Davidlohr Bueso787df632016-04-12 08:52:55 -0700466 ACQUIRE operations include LOCK operations and both smp_load_acquire()
467 and smp_cond_acquire() operations. The later builds the necessary ACQUIRE
468 semantics from relying on a control dependency and smp_rmb().
David Howells108b42b2006-03-31 16:00:29 +0100469
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100470 Memory operations that occur before an ACQUIRE operation may appear to
471 happen after it completes.
David Howells108b42b2006-03-31 16:00:29 +0100472
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100473 An ACQUIRE operation should almost always be paired with a RELEASE
474 operation.
David Howells108b42b2006-03-31 16:00:29 +0100475
476
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100477 (6) RELEASE operations.
David Howells108b42b2006-03-31 16:00:29 +0100478
479 This also acts as a one-way permeable barrier. It guarantees that all
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100480 memory operations before the RELEASE operation will appear to happen
481 before the RELEASE operation with respect to the other components of the
482 system. RELEASE operations include UNLOCK operations and
483 smp_store_release() operations.
David Howells108b42b2006-03-31 16:00:29 +0100484
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100485 Memory operations that occur after a RELEASE operation may appear to
David Howells108b42b2006-03-31 16:00:29 +0100486 happen before it completes.
487
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100488 The use of ACQUIRE and RELEASE operations generally precludes the need
489 for other sorts of memory barrier (but note the exceptions mentioned in
490 the subsection "MMIO write barrier"). In addition, a RELEASE+ACQUIRE
491 pair is -not- guaranteed to act as a full memory barrier. However, after
492 an ACQUIRE on a given variable, all memory accesses preceding any prior
493 RELEASE on that same variable are guaranteed to be visible. In other
494 words, within a given variable's critical section, all accesses of all
495 previous critical sections for that variable are guaranteed to have
496 completed.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -0800497
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100498 This means that ACQUIRE acts as a minimal "acquire" operation and
499 RELEASE acts as a minimal "release" operation.
David Howells108b42b2006-03-31 16:00:29 +0100500
Peter Zijlstra706eeb32017-06-12 14:50:27 +0200501A subset of the atomic operations described in atomic_t.txt have ACQUIRE and
502RELEASE variants in addition to fully-ordered and relaxed (no barrier
503semantics) definitions. For compound atomics performing both a load and a
504store, ACQUIRE semantics apply only to the load and RELEASE semantics apply
505only to the store portion of the operation.
David Howells108b42b2006-03-31 16:00:29 +0100506
507Memory barriers are only required where there's a possibility of interaction
508between two CPUs or between a CPU and a device. If it can be guaranteed that
509there won't be any such interaction in any particular piece of code, then
510memory barriers are unnecessary in that piece of code.
511
512
513Note that these are the _minimum_ guarantees. Different architectures may give
514more substantial guarantees, but they may _not_ be relied upon outside of arch
515specific code.
516
517
518WHAT MAY NOT BE ASSUMED ABOUT MEMORY BARRIERS?
519----------------------------------------------
520
521There are certain things that the Linux kernel memory barriers do not guarantee:
522
523 (*) There is no guarantee that any of the memory accesses specified before a
524 memory barrier will be _complete_ by the completion of a memory barrier
525 instruction; the barrier can be considered to draw a line in that CPU's
526 access queue that accesses of the appropriate type may not cross.
527
528 (*) There is no guarantee that issuing a memory barrier on one CPU will have
529 any direct effect on another CPU or any other hardware in the system. The
530 indirect effect will be the order in which the second CPU sees the effects
531 of the first CPU's accesses occur, but see the next point:
532
David Howells6bc39272006-06-25 05:49:22 -0700533 (*) There is no guarantee that a CPU will see the correct order of effects
David Howells108b42b2006-03-31 16:00:29 +0100534 from a second CPU's accesses, even _if_ the second CPU uses a memory
535 barrier, unless the first CPU _also_ uses a matching memory barrier (see
536 the subsection on "SMP Barrier Pairing").
537
538 (*) There is no guarantee that some intervening piece of off-the-CPU
539 hardware[*] will not reorder the memory accesses. CPU cache coherency
540 mechanisms should propagate the indirect effects of a memory barrier
541 between CPUs, but might not do so in order.
542
543 [*] For information on bus mastering DMA and coherency please read:
544
Randy Dunlap4b5ff462008-03-10 17:16:32 -0700545 Documentation/PCI/pci.txt
Paul Bolle395cf962011-08-15 02:02:26 +0200546 Documentation/DMA-API-HOWTO.txt
David Howells108b42b2006-03-31 16:00:29 +0100547 Documentation/DMA-API.txt
548
549
550DATA DEPENDENCY BARRIERS
551------------------------
552
553The usage requirements of data dependency barriers are a little subtle, and
554it's not always obvious that they're needed. To illustrate, consider the
555following sequence of events:
556
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800557 CPU 1 CPU 2
558 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700559 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100560 B = 4;
561 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700562 WRITE_ONCE(P, &B)
563 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800564 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100565
566There's a clear data dependency here, and it would seem that by the end of the
567sequence, Q must be either &A or &B, and that:
568
569 (Q == &A) implies (D == 1)
570 (Q == &B) implies (D == 4)
571
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700572But! CPU 2's perception of P may be updated _before_ its perception of B, thus
David Howells108b42b2006-03-31 16:00:29 +0100573leading to the following situation:
574
575 (Q == &B) and (D == 2) ????
576
577Whilst this may seem like a failure of coherency or causality maintenance, it
578isn't, and this behaviour can be observed on certain real CPUs (such as the DEC
579Alpha).
580
David Howells2b948952006-06-25 05:48:49 -0700581To deal with this, a data dependency barrier or better must be inserted
582between the address load and the data load:
David Howells108b42b2006-03-31 16:00:29 +0100583
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800584 CPU 1 CPU 2
585 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700586 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100587 B = 4;
588 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700589 WRITE_ONCE(P, &B);
590 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800591 <data dependency barrier>
592 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100593
594This enforces the occurrence of one of the two implications, and prevents the
595third possibility from arising.
596
Paul E. McKenney92a84dd2016-01-14 14:17:04 -0800597
David Howells108b42b2006-03-31 16:00:29 +0100598[!] Note that this extremely counterintuitive situation arises most easily on
599machines with split caches, so that, for example, one cache bank processes
600even-numbered cache lines and the other bank processes odd-numbered cache
601lines. The pointer P might be stored in an odd-numbered cache line, and the
602variable B might be stored in an even-numbered cache line. Then, if the
603even-numbered bank of the reading CPU's cache is extremely busy while the
604odd-numbered bank is idle, one can see the new value of the pointer P (&B),
David Howells6bc39272006-06-25 05:49:22 -0700605but the old value of the variable B (2).
David Howells108b42b2006-03-31 16:00:29 +0100606
607
Paul E. McKenney66ce3a42017-06-30 16:18:28 -0700608A data-dependency barrier is not required to order dependent writes
609because the CPUs that the Linux kernel supports don't do writes
610until they are certain (1) that the write will actually happen, (2)
611of the location of the write, and (3) of the value to be written.
612But please carefully read the "CONTROL DEPENDENCIES" section and the
613Documentation/RCU/rcu_dereference.txt file: The compiler can and does
614break dependencies in a great many highly creative ways.
615
616 CPU 1 CPU 2
617 =============== ===============
618 { A == 1, B == 2, C = 3, P == &A, Q == &C }
619 B = 4;
620 <write barrier>
621 WRITE_ONCE(P, &B);
622 Q = READ_ONCE(P);
623 WRITE_ONCE(*Q, 5);
624
625Therefore, no data-dependency barrier is required to order the read into
626Q with the store into *Q. In other words, this outcome is prohibited,
627even without a data-dependency barrier:
628
629 (Q == &B) && (B == 4)
630
631Please note that this pattern should be rare. After all, the whole point
632of dependency ordering is to -prevent- writes to the data structure, along
633with the expensive cache misses associated with those writes. This pattern
634can be used to record rare error conditions and the like, and the CPUs'
635naturally occurring ordering prevents such records from being lost.
636
637
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800638The data dependency barrier is very important to the RCU system,
639for example. See rcu_assign_pointer() and rcu_dereference() in
640include/linux/rcupdate.h. This permits the current target of an RCU'd
641pointer to be replaced with a new modified target, without the replacement
642target appearing to be incompletely initialised.
David Howells108b42b2006-03-31 16:00:29 +0100643
644See also the subsection on "Cache Coherency" for a more thorough example.
645
646
647CONTROL DEPENDENCIES
648--------------------
649
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800650Control dependencies can be a bit tricky because current compilers do
651not understand them. The purpose of this section is to help you prevent
652the compiler's ignorance from breaking your code.
653
Paul E. McKenneyff382812015-02-17 10:00:06 -0800654A load-load control dependency requires a full read memory barrier, not
655simply a data dependency barrier to make it work correctly. Consider the
656following bit of code:
David Howells108b42b2006-03-31 16:00:29 +0100657
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700658 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800659 if (q) {
660 <data dependency barrier> /* BUG: No data dependency!!! */
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700661 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700662 }
David Howells108b42b2006-03-31 16:00:29 +0100663
664This will not have the desired effect because there is no actual data
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800665dependency, but rather a control dependency that the CPU may short-circuit
666by attempting to predict the outcome in advance, so that other CPUs see
667the load from b as having happened before the load from a. In such a
668case what's actually required is:
David Howells108b42b2006-03-31 16:00:29 +0100669
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700670 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800671 if (q) {
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700672 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700673 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700674 }
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800675
676However, stores are not speculated. This means that ordering -is- provided
Paul E. McKenneyff382812015-02-17 10:00:06 -0800677for load-store control dependencies, as in the following example:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800678
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800679 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700680 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800681 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800682 }
683
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800684Control dependencies pair normally with other types of barriers.
685That said, please note that neither READ_ONCE() nor WRITE_ONCE()
686are optional! Without the READ_ONCE(), the compiler might combine the
687load from 'a' with other loads from 'a'. Without the WRITE_ONCE(),
688the compiler might combine the store to 'b' with other stores to 'b'.
689Either can result in highly counterintuitive effects on ordering.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800690
691Worse yet, if the compiler is able to prove (say) that the value of
692variable 'a' is always non-zero, it would be well within its rights
693to optimize the original example by eliminating the "if" statement
694as follows:
695
696 q = a;
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800697 b = 1; /* BUG: Compiler and CPU can both reorder!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800698
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800699So don't leave out the READ_ONCE().
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700700
701It is tempting to try to enforce ordering on identical stores on both
702branches of the "if" statement as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800703
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800704 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800705 if (q) {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800706 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800707 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800708 do_something();
709 } else {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800710 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800711 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800712 do_something_else();
713 }
714
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700715Unfortunately, current compilers will transform this as follows at high
716optimization levels:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800717
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800718 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700719 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800720 WRITE_ONCE(b, 1); /* BUG: No ordering vs. load from a!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800721 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800722 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800723 do_something();
724 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800725 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800726 do_something_else();
727 }
728
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700729Now there is no conditional between the load from 'a' and the store to
730'b', which means that the CPU is within its rights to reorder them:
731The conditional is absolutely required, and must be present in the
732assembly code even after all compiler optimizations have been applied.
733Therefore, if you need ordering in this example, you need explicit
734memory barriers, for example, smp_store_release():
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800735
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700736 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700737 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800738 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800739 do_something();
740 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800741 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800742 do_something_else();
743 }
744
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700745In contrast, without explicit memory barriers, two-legged-if control
746ordering is guaranteed only when the stores differ, for example:
747
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800748 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700749 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800750 WRITE_ONCE(b, 1);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700751 do_something();
752 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800753 WRITE_ONCE(b, 2);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700754 do_something_else();
755 }
756
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800757The initial READ_ONCE() is still required to prevent the compiler from
758proving the value of 'a'.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800759
760In addition, you need to be careful what you do with the local variable 'q',
761otherwise the compiler might be able to guess the value and again remove
762the needed conditional. For example:
763
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800764 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800765 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800766 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800767 do_something();
768 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800769 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800770 do_something_else();
771 }
772
773If MAX is defined to be 1, then the compiler knows that (q % MAX) is
774equal to zero, in which case the compiler is within its rights to
775transform the above code into the following:
776
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800777 q = READ_ONCE(a);
pierre Kuob26cfc42017-04-07 14:37:36 +0800778 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800779 do_something_else();
780
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700781Given this transformation, the CPU is not required to respect the ordering
782between the load from variable 'a' and the store to variable 'b'. It is
783tempting to add a barrier(), but this does not help. The conditional
784is gone, and the barrier won't bring it back. Therefore, if you are
785relying on this ordering, you should make sure that MAX is greater than
786one, perhaps as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800787
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800788 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800789 BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
790 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800791 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800792 do_something();
793 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800794 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800795 do_something_else();
796 }
797
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700798Please note once again that the stores to 'b' differ. If they were
799identical, as noted earlier, the compiler could pull this store outside
800of the 'if' statement.
801
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700802You must also be careful not to rely too much on boolean short-circuit
803evaluation. Consider this example:
804
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800805 q = READ_ONCE(a);
Paul E. McKenney57aecae2015-05-18 18:27:42 -0700806 if (q || 1 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700807 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700808
Paul E. McKenney5af46922015-04-25 12:48:29 -0700809Because the first condition cannot fault and the second condition is
810always true, the compiler can transform this example as following,
811defeating control dependency:
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700812
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800813 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700814 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700815
816This example underscores the need to ensure that the compiler cannot
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700817out-guess your code. More generally, although READ_ONCE() does force
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700818the compiler to actually emit code for a given load, it does not force
819the compiler to use the results.
820
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700821In addition, control dependencies apply only to the then-clause and
822else-clause of the if-statement in question. In particular, it does
823not necessarily apply to code following the if-statement:
824
825 q = READ_ONCE(a);
826 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800827 WRITE_ONCE(b, 1);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700828 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800829 WRITE_ONCE(b, 2);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700830 }
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800831 WRITE_ONCE(c, 1); /* BUG: No ordering against the read from 'a'. */
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700832
833It is tempting to argue that there in fact is ordering because the
834compiler cannot reorder volatile accesses and also cannot reorder
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800835the writes to 'b' with the condition. Unfortunately for this line
836of reasoning, the compiler might compile the two writes to 'b' as
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700837conditional-move instructions, as in this fanciful pseudo-assembly
838language:
839
840 ld r1,a
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700841 cmp r1,$0
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800842 cmov,ne r4,$1
843 cmov,eq r4,$2
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700844 st r4,b
845 st $1,c
846
847A weakly ordered CPU would have no dependency of any sort between the load
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800848from 'a' and the store to 'c'. The control dependencies would extend
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700849only to the pair of cmov instructions and the store depending on them.
850In short, control dependencies apply only to the stores in the then-clause
851and else-clause of the if-statement in question (including functions
852invoked by those two clauses), not to code following that if-statement.
853
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800854Finally, control dependencies do -not- provide transitivity. This is
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700855demonstrated by two related examples, with the initial values of
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800856'x' and 'y' both being zero:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800857
858 CPU 0 CPU 1
Paul E. McKenney5af46922015-04-25 12:48:29 -0700859 ======================= =======================
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800860 r1 = READ_ONCE(x); r2 = READ_ONCE(y);
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700861 if (r1 > 0) if (r2 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700862 WRITE_ONCE(y, 1); WRITE_ONCE(x, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800863
864 assert(!(r1 == 1 && r2 == 1));
865
866The above two-CPU example will never trigger the assert(). However,
867if control dependencies guaranteed transitivity (which they do not),
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700868then adding the following CPU would guarantee a related assertion:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800869
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700870 CPU 2
871 =====================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700872 WRITE_ONCE(x, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800873
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700874 assert(!(r1 == 2 && r2 == 1 && x == 2)); /* FAILS!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800875
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700876But because control dependencies do -not- provide transitivity, the above
877assertion can fail after the combined three-CPU example completes. If you
878need the three-CPU example to provide ordering, you will need smp_mb()
879between the loads and stores in the CPU 0 and CPU 1 code fragments,
Paul E. McKenney5af46922015-04-25 12:48:29 -0700880that is, just before or just after the "if" statements. Furthermore,
881the original two-CPU example is very fragile and should be avoided.
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700882
883These two examples are the LB and WWC litmus tests from this paper:
884http://www.cl.cam.ac.uk/users/pes20/ppc-supplemental/test6.pdf and this
885site: https://www.cl.cam.ac.uk/~pes20/ppcmem/index.html.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800886
887In summary:
888
889 (*) Control dependencies can order prior loads against later stores.
890 However, they do -not- guarantee any other sort of ordering:
891 Not prior loads against later loads, nor prior stores against
892 later anything. If you need these other forms of ordering,
Davidlohr Buesod87510c2014-12-28 01:11:16 -0800893 use smp_rmb(), smp_wmb(), or, in the case of prior stores and
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800894 later loads, smp_mb().
895
Paul E. McKenney7817b792015-12-29 16:23:18 -0800896 (*) If both legs of the "if" statement begin with identical stores to
897 the same variable, then those stores must be ordered, either by
898 preceding both of them with smp_mb() or by using smp_store_release()
899 to carry out the stores. Please note that it is -not- sufficient
Paul E. McKenneya5052652016-04-12 08:52:49 -0700900 to use barrier() at beginning of each leg of the "if" statement
901 because, as shown by the example above, optimizing compilers can
902 destroy the control dependency while respecting the letter of the
903 barrier() law.
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800904
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800905 (*) Control dependencies require at least one run-time conditional
Paul E. McKenney586dd562014-02-11 12:28:06 -0800906 between the prior load and the subsequent store, and this
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700907 conditional must involve the prior load. If the compiler is able
908 to optimize the conditional away, it will have also optimized
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800909 away the ordering. Careful use of READ_ONCE() and WRITE_ONCE()
910 can help to preserve the needed conditional.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800911
912 (*) Control dependencies require that the compiler avoid reordering the
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800913 dependency into nonexistence. Careful use of READ_ONCE() or
914 atomic{,64}_read() can help to preserve your control dependency.
Paul E. McKenney895f5542016-01-06 14:23:03 -0800915 Please see the COMPILER BARRIER section for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800916
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700917 (*) Control dependencies apply only to the then-clause and else-clause
918 of the if-statement containing the control dependency, including
919 any functions that these two clauses call. Control dependencies
920 do -not- apply to code following the if-statement containing the
921 control dependency.
922
Paul E. McKenneyff382812015-02-17 10:00:06 -0800923 (*) Control dependencies pair normally with other types of barriers.
924
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800925 (*) Control dependencies do -not- provide transitivity. If you
926 need transitivity, use smp_mb().
David Howells108b42b2006-03-31 16:00:29 +0100927
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800928 (*) Compilers do not understand control dependencies. It is therefore
929 your job to ensure that they do not break your code.
930
David Howells108b42b2006-03-31 16:00:29 +0100931
932SMP BARRIER PAIRING
933-------------------
934
935When dealing with CPU-CPU interactions, certain types of memory barrier should
936always be paired. A lack of appropriate pairing is almost certainly an error.
937
Paul E. McKenneyff382812015-02-17 10:00:06 -0800938General barriers pair with each other, though they also pair with most
939other types of barriers, albeit without transitivity. An acquire barrier
940pairs with a release barrier, but both may also pair with other barriers,
941including of course general barriers. A write barrier pairs with a data
942dependency barrier, a control dependency, an acquire barrier, a release
943barrier, a read barrier, or a general barrier. Similarly a read barrier,
944control dependency, or a data dependency barrier pairs with a write
945barrier, an acquire barrier, a release barrier, or a general barrier:
David Howells108b42b2006-03-31 16:00:29 +0100946
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800947 CPU 1 CPU 2
948 =============== ===============
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700949 WRITE_ONCE(a, 1);
David Howells108b42b2006-03-31 16:00:29 +0100950 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700951 WRITE_ONCE(b, 2); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800952 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700953 y = READ_ONCE(a);
David Howells108b42b2006-03-31 16:00:29 +0100954
955Or:
956
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800957 CPU 1 CPU 2
958 =============== ===============================
David Howells108b42b2006-03-31 16:00:29 +0100959 a = 1;
960 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700961 WRITE_ONCE(b, &a); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800962 <data dependency barrier>
963 y = *x;
David Howells108b42b2006-03-31 16:00:29 +0100964
Paul E. McKenneyff382812015-02-17 10:00:06 -0800965Or even:
966
967 CPU 1 CPU 2
968 =============== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700969 r1 = READ_ONCE(y);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800970 <general barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700971 WRITE_ONCE(y, 1); if (r2 = READ_ONCE(x)) {
Paul E. McKenneyff382812015-02-17 10:00:06 -0800972 <implicit control dependency>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700973 WRITE_ONCE(y, 1);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800974 }
975
976 assert(r1 == 0 || r2 == 0);
977
David Howells108b42b2006-03-31 16:00:29 +0100978Basically, the read barrier always has to be there, even though it can be of
979the "weaker" type.
980
David Howells670bd952006-06-10 09:54:12 -0700981[!] Note that the stores before the write barrier would normally be expected to
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700982match the loads after the read barrier or the data dependency barrier, and vice
David Howells670bd952006-06-10 09:54:12 -0700983versa:
984
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800985 CPU 1 CPU 2
986 =================== ===================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700987 WRITE_ONCE(a, 1); }---- --->{ v = READ_ONCE(c);
988 WRITE_ONCE(b, 2); } \ / { w = READ_ONCE(d);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800989 <write barrier> \ <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700990 WRITE_ONCE(c, 3); } / \ { x = READ_ONCE(a);
991 WRITE_ONCE(d, 4); }---- --->{ y = READ_ONCE(b);
David Howells670bd952006-06-10 09:54:12 -0700992
David Howells108b42b2006-03-31 16:00:29 +0100993
994EXAMPLES OF MEMORY BARRIER SEQUENCES
995------------------------------------
996
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700997Firstly, write barriers act as partial orderings on store operations.
David Howells108b42b2006-03-31 16:00:29 +0100998Consider the following sequence of events:
999
1000 CPU 1
1001 =======================
1002 STORE A = 1
1003 STORE B = 2
1004 STORE C = 3
1005 <write barrier>
1006 STORE D = 4
1007 STORE E = 5
1008
1009This sequence of events is committed to the memory coherence system in an order
1010that the rest of the system might perceive as the unordered set of { STORE A,
Adrian Bunk80f72282006-06-30 18:27:16 +02001011STORE B, STORE C } all occurring before the unordered set of { STORE D, STORE E
David Howells108b42b2006-03-31 16:00:29 +01001012}:
1013
1014 +-------+ : :
1015 | | +------+
1016 | |------>| C=3 | } /\
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001017 | | : +------+ }----- \ -----> Events perceptible to
1018 | | : | A=1 | } \/ the rest of the system
David Howells108b42b2006-03-31 16:00:29 +01001019 | | : +------+ }
1020 | CPU 1 | : | B=2 | }
1021 | | +------+ }
1022 | | wwwwwwwwwwwwwwww } <--- At this point the write barrier
1023 | | +------+ } requires all stores prior to the
1024 | | : | E=5 | } barrier to be committed before
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001025 | | : +------+ } further stores may take place
David Howells108b42b2006-03-31 16:00:29 +01001026 | |------>| D=4 | }
1027 | | +------+
1028 +-------+ : :
1029 |
David Howells670bd952006-06-10 09:54:12 -07001030 | Sequence in which stores are committed to the
1031 | memory system by CPU 1
David Howells108b42b2006-03-31 16:00:29 +01001032 V
1033
1034
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001035Secondly, data dependency barriers act as partial orderings on data-dependent
David Howells108b42b2006-03-31 16:00:29 +01001036loads. Consider the following sequence of events:
1037
1038 CPU 1 CPU 2
1039 ======================= =======================
David Howellsc14038c2006-04-10 22:54:24 -07001040 { B = 7; X = 9; Y = 8; C = &Y }
David Howells108b42b2006-03-31 16:00:29 +01001041 STORE A = 1
1042 STORE B = 2
1043 <write barrier>
1044 STORE C = &B LOAD X
1045 STORE D = 4 LOAD C (gets &B)
1046 LOAD *C (reads B)
1047
1048Without intervention, CPU 2 may perceive the events on CPU 1 in some
1049effectively random order, despite the write barrier issued by CPU 1:
1050
1051 +-------+ : : : :
1052 | | +------+ +-------+ | Sequence of update
1053 | |------>| B=2 |----- --->| Y->8 | | of perception on
1054 | | : +------+ \ +-------+ | CPU 2
1055 | CPU 1 | : | A=1 | \ --->| C->&Y | V
1056 | | +------+ | +-------+
1057 | | wwwwwwwwwwwwwwww | : :
1058 | | +------+ | : :
1059 | | : | C=&B |--- | : : +-------+
1060 | | : +------+ \ | +-------+ | |
1061 | |------>| D=4 | ----------->| C->&B |------>| |
1062 | | +------+ | +-------+ | |
1063 +-------+ : : | : : | |
1064 | : : | |
1065 | : : | CPU 2 |
1066 | +-------+ | |
1067 Apparently incorrect ---> | | B->7 |------>| |
1068 perception of B (!) | +-------+ | |
1069 | : : | |
1070 | +-------+ | |
1071 The load of X holds ---> \ | X->9 |------>| |
1072 up the maintenance \ +-------+ | |
1073 of coherence of B ----->| B->2 | +-------+
1074 +-------+
1075 : :
1076
1077
1078In the above example, CPU 2 perceives that B is 7, despite the load of *C
Paolo Ornati670e9f32006-10-03 22:57:56 +02001079(which would be B) coming after the LOAD of C.
David Howells108b42b2006-03-31 16:00:29 +01001080
1081If, however, a data dependency barrier were to be placed between the load of C
David Howellsc14038c2006-04-10 22:54:24 -07001082and the load of *C (ie: B) on CPU 2:
1083
1084 CPU 1 CPU 2
1085 ======================= =======================
1086 { B = 7; X = 9; Y = 8; C = &Y }
1087 STORE A = 1
1088 STORE B = 2
1089 <write barrier>
1090 STORE C = &B LOAD X
1091 STORE D = 4 LOAD C (gets &B)
1092 <data dependency barrier>
1093 LOAD *C (reads B)
1094
1095then the following will occur:
David Howells108b42b2006-03-31 16:00:29 +01001096
1097 +-------+ : : : :
1098 | | +------+ +-------+
1099 | |------>| B=2 |----- --->| Y->8 |
1100 | | : +------+ \ +-------+
1101 | CPU 1 | : | A=1 | \ --->| C->&Y |
1102 | | +------+ | +-------+
1103 | | wwwwwwwwwwwwwwww | : :
1104 | | +------+ | : :
1105 | | : | C=&B |--- | : : +-------+
1106 | | : +------+ \ | +-------+ | |
1107 | |------>| D=4 | ----------->| C->&B |------>| |
1108 | | +------+ | +-------+ | |
1109 +-------+ : : | : : | |
1110 | : : | |
1111 | : : | CPU 2 |
1112 | +-------+ | |
David Howells670bd952006-06-10 09:54:12 -07001113 | | X->9 |------>| |
1114 | +-------+ | |
1115 Makes sure all effects ---> \ ddddddddddddddddd | |
1116 prior to the store of C \ +-------+ | |
1117 are perceptible to ----->| B->2 |------>| |
1118 subsequent loads +-------+ | |
David Howells108b42b2006-03-31 16:00:29 +01001119 : : +-------+
1120
1121
1122And thirdly, a read barrier acts as a partial order on loads. Consider the
1123following sequence of events:
1124
1125 CPU 1 CPU 2
1126 ======================= =======================
David Howells670bd952006-06-10 09:54:12 -07001127 { A = 0, B = 9 }
David Howells108b42b2006-03-31 16:00:29 +01001128 STORE A=1
David Howells108b42b2006-03-31 16:00:29 +01001129 <write barrier>
David Howells670bd952006-06-10 09:54:12 -07001130 STORE B=2
David Howells108b42b2006-03-31 16:00:29 +01001131 LOAD B
David Howells670bd952006-06-10 09:54:12 -07001132 LOAD A
David Howells108b42b2006-03-31 16:00:29 +01001133
1134Without intervention, CPU 2 may then choose to perceive the events on CPU 1 in
1135some effectively random order, despite the write barrier issued by CPU 1:
1136
David Howells670bd952006-06-10 09:54:12 -07001137 +-------+ : : : :
1138 | | +------+ +-------+
1139 | |------>| A=1 |------ --->| A->0 |
1140 | | +------+ \ +-------+
1141 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1142 | | +------+ | +-------+
1143 | |------>| B=2 |--- | : :
1144 | | +------+ \ | : : +-------+
1145 +-------+ : : \ | +-------+ | |
1146 ---------->| B->2 |------>| |
1147 | +-------+ | CPU 2 |
1148 | | A->0 |------>| |
1149 | +-------+ | |
1150 | : : +-------+
1151 \ : :
1152 \ +-------+
1153 ---->| A->1 |
1154 +-------+
1155 : :
David Howells108b42b2006-03-31 16:00:29 +01001156
1157
David Howells6bc39272006-06-25 05:49:22 -07001158If, however, a read barrier were to be placed between the load of B and the
David Howells670bd952006-06-10 09:54:12 -07001159load of A on CPU 2:
David Howells108b42b2006-03-31 16:00:29 +01001160
David Howells670bd952006-06-10 09:54:12 -07001161 CPU 1 CPU 2
1162 ======================= =======================
1163 { A = 0, B = 9 }
1164 STORE A=1
1165 <write barrier>
1166 STORE B=2
1167 LOAD B
1168 <read barrier>
1169 LOAD A
1170
1171then the partial ordering imposed by CPU 1 will be perceived correctly by CPU
11722:
1173
1174 +-------+ : : : :
1175 | | +------+ +-------+
1176 | |------>| A=1 |------ --->| A->0 |
1177 | | +------+ \ +-------+
1178 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1179 | | +------+ | +-------+
1180 | |------>| B=2 |--- | : :
1181 | | +------+ \ | : : +-------+
1182 +-------+ : : \ | +-------+ | |
1183 ---------->| B->2 |------>| |
1184 | +-------+ | CPU 2 |
1185 | : : | |
1186 | : : | |
1187 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1188 barrier causes all effects \ +-------+ | |
1189 prior to the storage of B ---->| A->1 |------>| |
1190 to be perceptible to CPU 2 +-------+ | |
1191 : : +-------+
1192
1193
1194To illustrate this more completely, consider what could happen if the code
1195contained a load of A either side of the read barrier:
1196
1197 CPU 1 CPU 2
1198 ======================= =======================
1199 { A = 0, B = 9 }
1200 STORE A=1
1201 <write barrier>
1202 STORE B=2
1203 LOAD B
1204 LOAD A [first load of A]
1205 <read barrier>
1206 LOAD A [second load of A]
1207
1208Even though the two loads of A both occur after the load of B, they may both
1209come up with different values:
1210
1211 +-------+ : : : :
1212 | | +------+ +-------+
1213 | |------>| A=1 |------ --->| A->0 |
1214 | | +------+ \ +-------+
1215 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1216 | | +------+ | +-------+
1217 | |------>| B=2 |--- | : :
1218 | | +------+ \ | : : +-------+
1219 +-------+ : : \ | +-------+ | |
1220 ---------->| B->2 |------>| |
1221 | +-------+ | CPU 2 |
1222 | : : | |
1223 | : : | |
1224 | +-------+ | |
1225 | | A->0 |------>| 1st |
1226 | +-------+ | |
1227 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1228 barrier causes all effects \ +-------+ | |
1229 prior to the storage of B ---->| A->1 |------>| 2nd |
1230 to be perceptible to CPU 2 +-------+ | |
1231 : : +-------+
1232
1233
1234But it may be that the update to A from CPU 1 becomes perceptible to CPU 2
1235before the read barrier completes anyway:
1236
1237 +-------+ : : : :
1238 | | +------+ +-------+
1239 | |------>| A=1 |------ --->| A->0 |
1240 | | +------+ \ +-------+
1241 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1242 | | +------+ | +-------+
1243 | |------>| B=2 |--- | : :
1244 | | +------+ \ | : : +-------+
1245 +-------+ : : \ | +-------+ | |
1246 ---------->| B->2 |------>| |
1247 | +-------+ | CPU 2 |
1248 | : : | |
1249 \ : : | |
1250 \ +-------+ | |
1251 ---->| A->1 |------>| 1st |
1252 +-------+ | |
1253 rrrrrrrrrrrrrrrrr | |
1254 +-------+ | |
1255 | A->1 |------>| 2nd |
1256 +-------+ | |
1257 : : +-------+
1258
1259
1260The guarantee is that the second load will always come up with A == 1 if the
1261load of B came up with B == 2. No such guarantee exists for the first load of
1262A; that may come up with either A == 0 or A == 1.
1263
1264
1265READ MEMORY BARRIERS VS LOAD SPECULATION
1266----------------------------------------
1267
1268Many CPUs speculate with loads: that is they see that they will need to load an
1269item from memory, and they find a time where they're not using the bus for any
1270other loads, and so do the load in advance - even though they haven't actually
1271got to that point in the instruction execution flow yet. This permits the
1272actual load instruction to potentially complete immediately because the CPU
1273already has the value to hand.
1274
1275It may turn out that the CPU didn't actually need the value - perhaps because a
1276branch circumvented the load - in which case it can discard the value or just
1277cache it for later use.
1278
1279Consider:
1280
Ingo Molnare0edc782013-11-22 11:24:53 +01001281 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001282 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001283 LOAD B
1284 DIVIDE } Divide instructions generally
1285 DIVIDE } take a long time to perform
1286 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001287
1288Which might appear as this:
1289
1290 : : +-------+
1291 +-------+ | |
1292 --->| B->2 |------>| |
1293 +-------+ | CPU 2 |
1294 : :DIVIDE | |
1295 +-------+ | |
1296 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1297 division speculates on the +-------+ ~ | |
1298 LOAD of A : : ~ | |
1299 : :DIVIDE | |
1300 : : ~ | |
1301 Once the divisions are complete --> : : ~-->| |
1302 the CPU can then perform the : : | |
1303 LOAD with immediate effect : : +-------+
1304
1305
1306Placing a read barrier or a data dependency barrier just before the second
1307load:
1308
Ingo Molnare0edc782013-11-22 11:24:53 +01001309 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001310 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001311 LOAD B
1312 DIVIDE
1313 DIVIDE
David Howells670bd952006-06-10 09:54:12 -07001314 <read barrier>
Ingo Molnare0edc782013-11-22 11:24:53 +01001315 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001316
1317will force any value speculatively obtained to be reconsidered to an extent
1318dependent on the type of barrier used. If there was no change made to the
1319speculated memory location, then the speculated value will just be used:
1320
1321 : : +-------+
1322 +-------+ | |
1323 --->| B->2 |------>| |
1324 +-------+ | CPU 2 |
1325 : :DIVIDE | |
1326 +-------+ | |
1327 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1328 division speculates on the +-------+ ~ | |
1329 LOAD of A : : ~ | |
1330 : :DIVIDE | |
1331 : : ~ | |
1332 : : ~ | |
1333 rrrrrrrrrrrrrrrr~ | |
1334 : : ~ | |
1335 : : ~-->| |
1336 : : | |
1337 : : +-------+
1338
1339
1340but if there was an update or an invalidation from another CPU pending, then
1341the speculation will be cancelled and the value reloaded:
1342
1343 : : +-------+
1344 +-------+ | |
1345 --->| B->2 |------>| |
1346 +-------+ | CPU 2 |
1347 : :DIVIDE | |
1348 +-------+ | |
1349 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1350 division speculates on the +-------+ ~ | |
1351 LOAD of A : : ~ | |
1352 : :DIVIDE | |
1353 : : ~ | |
1354 : : ~ | |
1355 rrrrrrrrrrrrrrrrr | |
1356 +-------+ | |
1357 The speculation is discarded ---> --->| A->1 |------>| |
1358 and an updated value is +-------+ | |
1359 retrieved : : +-------+
David Howells108b42b2006-03-31 16:00:29 +01001360
1361
Paul E. McKenney241e6662011-02-10 16:54:50 -08001362TRANSITIVITY
1363------------
1364
1365Transitivity is a deeply intuitive notion about ordering that is not
1366always provided by real computer systems. The following example
Paul E. McKenneyf36fe1e2016-02-15 14:50:36 -08001367demonstrates transitivity:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001368
1369 CPU 1 CPU 2 CPU 3
1370 ======================= ======================= =======================
1371 { X = 0, Y = 0 }
1372 STORE X=1 LOAD X STORE Y=1
1373 <general barrier> <general barrier>
1374 LOAD Y LOAD X
1375
1376Suppose that CPU 2's load from X returns 1 and its load from Y returns 0.
1377This indicates that CPU 2's load from X in some sense follows CPU 1's
1378store to X and that CPU 2's load from Y in some sense preceded CPU 3's
1379store to Y. The question is then "Can CPU 3's load from X return 0?"
1380
1381Because CPU 2's load from X in some sense came after CPU 1's store, it
1382is natural to expect that CPU 3's load from X must therefore return 1.
1383This expectation is an example of transitivity: if a load executing on
1384CPU A follows a load from the same variable executing on CPU B, then
1385CPU A's load must either return the same value that CPU B's load did,
1386or must return some later value.
1387
1388In the Linux kernel, use of general memory barriers guarantees
1389transitivity. Therefore, in the above example, if CPU 2's load from X
1390returns 1 and its load from Y returns 0, then CPU 3's load from X must
1391also return 1.
1392
1393However, transitivity is -not- guaranteed for read or write barriers.
1394For example, suppose that CPU 2's general barrier in the above example
1395is changed to a read barrier as shown below:
1396
1397 CPU 1 CPU 2 CPU 3
1398 ======================= ======================= =======================
1399 { X = 0, Y = 0 }
1400 STORE X=1 LOAD X STORE Y=1
1401 <read barrier> <general barrier>
1402 LOAD Y LOAD X
1403
1404This substitution destroys transitivity: in this example, it is perfectly
1405legal for CPU 2's load from X to return 1, its load from Y to return 0,
1406and CPU 3's load from X to return 0.
1407
1408The key point is that although CPU 2's read barrier orders its pair
1409of loads, it does not guarantee to order CPU 1's store. Therefore, if
1410this example runs on a system where CPUs 1 and 2 share a store buffer
1411or a level of cache, CPU 2 might have early access to CPU 1's writes.
1412General barriers are therefore required to ensure that all CPUs agree
1413on the combined order of CPU 1's and CPU 2's accesses.
1414
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001415General barriers provide "global transitivity", so that all CPUs will
1416agree on the order of operations. In contrast, a chain of release-acquire
1417pairs provides only "local transitivity", so that only those CPUs on
1418the chain are guaranteed to agree on the combined order of the accesses.
1419For example, switching to C code in deference to Herman Hollerith:
1420
1421 int u, v, x, y, z;
1422
1423 void cpu0(void)
1424 {
1425 r0 = smp_load_acquire(&x);
1426 WRITE_ONCE(u, 1);
1427 smp_store_release(&y, 1);
1428 }
1429
1430 void cpu1(void)
1431 {
1432 r1 = smp_load_acquire(&y);
1433 r4 = READ_ONCE(v);
1434 r5 = READ_ONCE(u);
1435 smp_store_release(&z, 1);
1436 }
1437
1438 void cpu2(void)
1439 {
1440 r2 = smp_load_acquire(&z);
1441 smp_store_release(&x, 1);
1442 }
1443
1444 void cpu3(void)
1445 {
1446 WRITE_ONCE(v, 1);
1447 smp_mb();
1448 r3 = READ_ONCE(u);
1449 }
1450
1451Because cpu0(), cpu1(), and cpu2() participate in a local transitive
1452chain of smp_store_release()/smp_load_acquire() pairs, the following
1453outcome is prohibited:
1454
1455 r0 == 1 && r1 == 1 && r2 == 1
1456
1457Furthermore, because of the release-acquire relationship between cpu0()
1458and cpu1(), cpu1() must see cpu0()'s writes, so that the following
1459outcome is prohibited:
1460
1461 r1 == 1 && r5 == 0
1462
1463However, the transitivity of release-acquire is local to the participating
1464CPUs and does not apply to cpu3(). Therefore, the following outcome
1465is possible:
1466
1467 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0
1468
Paul E. McKenney37ef0342016-01-25 22:12:34 -08001469As an aside, the following outcome is also possible:
1470
1471 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0 && r5 == 1
1472
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001473Although cpu0(), cpu1(), and cpu2() will see their respective reads and
1474writes in order, CPUs not involved in the release-acquire chain might
1475well disagree on the order. This disagreement stems from the fact that
1476the weak memory-barrier instructions used to implement smp_load_acquire()
1477and smp_store_release() are not required to order prior stores against
1478subsequent loads in all cases. This means that cpu3() can see cpu0()'s
1479store to u as happening -after- cpu1()'s load from v, even though
1480both cpu0() and cpu1() agree that these two operations occurred in the
1481intended order.
1482
1483However, please keep in mind that smp_load_acquire() is not magic.
1484In particular, it simply reads from its argument with ordering. It does
1485-not- ensure that any particular value will be read. Therefore, the
1486following outcome is possible:
1487
1488 r0 == 0 && r1 == 0 && r2 == 0 && r5 == 0
1489
1490Note that this outcome can happen even on a mythical sequentially
1491consistent system where nothing is ever reordered.
1492
1493To reiterate, if your code requires global transitivity, use general
1494barriers throughout.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001495
1496
David Howells108b42b2006-03-31 16:00:29 +01001497========================
1498EXPLICIT KERNEL BARRIERS
1499========================
1500
1501The Linux kernel has a variety of different barriers that act at different
1502levels:
1503
1504 (*) Compiler barrier.
1505
1506 (*) CPU memory barriers.
1507
1508 (*) MMIO write barrier.
1509
1510
1511COMPILER BARRIER
1512----------------
1513
1514The Linux kernel has an explicit compiler barrier function that prevents the
1515compiler from moving the memory accesses either side of it to the other side:
1516
1517 barrier();
1518
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001519This is a general barrier -- there are no read-read or write-write
1520variants of barrier(). However, READ_ONCE() and WRITE_ONCE() can be
1521thought of as weak forms of barrier() that affect only the specific
1522accesses flagged by the READ_ONCE() or WRITE_ONCE().
David Howells108b42b2006-03-31 16:00:29 +01001523
Paul E. McKenney692118d2013-12-11 13:59:07 -08001524The barrier() function has the following effects:
1525
1526 (*) Prevents the compiler from reordering accesses following the
1527 barrier() to precede any accesses preceding the barrier().
1528 One example use for this property is to ease communication between
1529 interrupt-handler code and the code that was interrupted.
1530
1531 (*) Within a loop, forces the compiler to load the variables used
1532 in that loop's conditional on each pass through that loop.
1533
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001534The READ_ONCE() and WRITE_ONCE() functions can prevent any number of
1535optimizations that, while perfectly safe in single-threaded code, can
1536be fatal in concurrent code. Here are some examples of these sorts
1537of optimizations:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001538
Paul E. McKenney449f7412014-01-02 15:03:50 -08001539 (*) The compiler is within its rights to reorder loads and stores
1540 to the same variable, and in some cases, the CPU is within its
1541 rights to reorder loads to the same variable. This means that
1542 the following code:
1543
1544 a[0] = x;
1545 a[1] = x;
1546
1547 Might result in an older value of x stored in a[1] than in a[0].
1548 Prevent both the compiler and the CPU from doing this as follows:
1549
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001550 a[0] = READ_ONCE(x);
1551 a[1] = READ_ONCE(x);
Paul E. McKenney449f7412014-01-02 15:03:50 -08001552
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001553 In short, READ_ONCE() and WRITE_ONCE() provide cache coherence for
1554 accesses from multiple CPUs to a single variable.
Paul E. McKenney449f7412014-01-02 15:03:50 -08001555
Paul E. McKenney692118d2013-12-11 13:59:07 -08001556 (*) The compiler is within its rights to merge successive loads from
1557 the same variable. Such merging can cause the compiler to "optimize"
1558 the following code:
1559
1560 while (tmp = a)
1561 do_something_with(tmp);
1562
1563 into the following code, which, although in some sense legitimate
1564 for single-threaded code, is almost certainly not what the developer
1565 intended:
1566
1567 if (tmp = a)
1568 for (;;)
1569 do_something_with(tmp);
1570
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001571 Use READ_ONCE() to prevent the compiler from doing this to you:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001572
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001573 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001574 do_something_with(tmp);
1575
1576 (*) The compiler is within its rights to reload a variable, for example,
1577 in cases where high register pressure prevents the compiler from
1578 keeping all data of interest in registers. The compiler might
1579 therefore optimize the variable 'tmp' out of our previous example:
1580
1581 while (tmp = a)
1582 do_something_with(tmp);
1583
1584 This could result in the following code, which is perfectly safe in
1585 single-threaded code, but can be fatal in concurrent code:
1586
1587 while (a)
1588 do_something_with(a);
1589
1590 For example, the optimized version of this code could result in
1591 passing a zero to do_something_with() in the case where the variable
1592 a was modified by some other CPU between the "while" statement and
1593 the call to do_something_with().
1594
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001595 Again, use READ_ONCE() to prevent the compiler from doing this:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001596
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001597 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001598 do_something_with(tmp);
1599
1600 Note that if the compiler runs short of registers, it might save
1601 tmp onto the stack. The overhead of this saving and later restoring
1602 is why compilers reload variables. Doing so is perfectly safe for
1603 single-threaded code, so you need to tell the compiler about cases
1604 where it is not safe.
1605
1606 (*) The compiler is within its rights to omit a load entirely if it knows
1607 what the value will be. For example, if the compiler can prove that
1608 the value of variable 'a' is always zero, it can optimize this code:
1609
1610 while (tmp = a)
1611 do_something_with(tmp);
1612
1613 Into this:
1614
1615 do { } while (0);
1616
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001617 This transformation is a win for single-threaded code because it
1618 gets rid of a load and a branch. The problem is that the compiler
1619 will carry out its proof assuming that the current CPU is the only
1620 one updating variable 'a'. If variable 'a' is shared, then the
1621 compiler's proof will be erroneous. Use READ_ONCE() to tell the
1622 compiler that it doesn't know as much as it thinks it does:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001623
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001624 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001625 do_something_with(tmp);
1626
1627 But please note that the compiler is also closely watching what you
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001628 do with the value after the READ_ONCE(). For example, suppose you
Paul E. McKenney692118d2013-12-11 13:59:07 -08001629 do the following and MAX is a preprocessor macro with the value 1:
1630
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001631 while ((tmp = READ_ONCE(a)) % MAX)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001632 do_something_with(tmp);
1633
1634 Then the compiler knows that the result of the "%" operator applied
1635 to MAX will always be zero, again allowing the compiler to optimize
1636 the code into near-nonexistence. (It will still load from the
1637 variable 'a'.)
1638
1639 (*) Similarly, the compiler is within its rights to omit a store entirely
1640 if it knows that the variable already has the value being stored.
1641 Again, the compiler assumes that the current CPU is the only one
1642 storing into the variable, which can cause the compiler to do the
1643 wrong thing for shared variables. For example, suppose you have
1644 the following:
1645
1646 a = 0;
SeongJae Park65f95ff2016-02-22 08:28:29 -08001647 ... Code that does not store to variable a ...
Paul E. McKenney692118d2013-12-11 13:59:07 -08001648 a = 0;
1649
1650 The compiler sees that the value of variable 'a' is already zero, so
1651 it might well omit the second store. This would come as a fatal
1652 surprise if some other CPU might have stored to variable 'a' in the
1653 meantime.
1654
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001655 Use WRITE_ONCE() to prevent the compiler from making this sort of
Paul E. McKenney692118d2013-12-11 13:59:07 -08001656 wrong guess:
1657
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001658 WRITE_ONCE(a, 0);
SeongJae Park65f95ff2016-02-22 08:28:29 -08001659 ... Code that does not store to variable a ...
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001660 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001661
1662 (*) The compiler is within its rights to reorder memory accesses unless
1663 you tell it not to. For example, consider the following interaction
1664 between process-level code and an interrupt handler:
1665
1666 void process_level(void)
1667 {
1668 msg = get_message();
1669 flag = true;
1670 }
1671
1672 void interrupt_handler(void)
1673 {
1674 if (flag)
1675 process_message(msg);
1676 }
1677
Masanari Iidadf5cbb22014-03-21 10:04:30 +09001678 There is nothing to prevent the compiler from transforming
Paul E. McKenney692118d2013-12-11 13:59:07 -08001679 process_level() to the following, in fact, this might well be a
1680 win for single-threaded code:
1681
1682 void process_level(void)
1683 {
1684 flag = true;
1685 msg = get_message();
1686 }
1687
1688 If the interrupt occurs between these two statement, then
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001689 interrupt_handler() might be passed a garbled msg. Use WRITE_ONCE()
Paul E. McKenney692118d2013-12-11 13:59:07 -08001690 to prevent this as follows:
1691
1692 void process_level(void)
1693 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001694 WRITE_ONCE(msg, get_message());
1695 WRITE_ONCE(flag, true);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001696 }
1697
1698 void interrupt_handler(void)
1699 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001700 if (READ_ONCE(flag))
1701 process_message(READ_ONCE(msg));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001702 }
1703
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001704 Note that the READ_ONCE() and WRITE_ONCE() wrappers in
1705 interrupt_handler() are needed if this interrupt handler can itself
1706 be interrupted by something that also accesses 'flag' and 'msg',
1707 for example, a nested interrupt or an NMI. Otherwise, READ_ONCE()
1708 and WRITE_ONCE() are not needed in interrupt_handler() other than
1709 for documentation purposes. (Note also that nested interrupts
1710 do not typically occur in modern Linux kernels, in fact, if an
1711 interrupt handler returns with interrupts enabled, you will get a
1712 WARN_ONCE() splat.)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001713
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001714 You should assume that the compiler can move READ_ONCE() and
1715 WRITE_ONCE() past code not containing READ_ONCE(), WRITE_ONCE(),
1716 barrier(), or similar primitives.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001717
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001718 This effect could also be achieved using barrier(), but READ_ONCE()
1719 and WRITE_ONCE() are more selective: With READ_ONCE() and
1720 WRITE_ONCE(), the compiler need only forget the contents of the
1721 indicated memory locations, while with barrier() the compiler must
1722 discard the value of all memory locations that it has currented
1723 cached in any machine registers. Of course, the compiler must also
1724 respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
1725 though the CPU of course need not do so.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001726
1727 (*) The compiler is within its rights to invent stores to a variable,
1728 as in the following example:
1729
1730 if (a)
1731 b = a;
1732 else
1733 b = 42;
1734
1735 The compiler might save a branch by optimizing this as follows:
1736
1737 b = 42;
1738 if (a)
1739 b = a;
1740
1741 In single-threaded code, this is not only safe, but also saves
1742 a branch. Unfortunately, in concurrent code, this optimization
1743 could cause some other CPU to see a spurious value of 42 -- even
1744 if variable 'a' was never zero -- when loading variable 'b'.
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001745 Use WRITE_ONCE() to prevent this as follows:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001746
1747 if (a)
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001748 WRITE_ONCE(b, a);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001749 else
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001750 WRITE_ONCE(b, 42);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001751
1752 The compiler can also invent loads. These are usually less
1753 damaging, but they can result in cache-line bouncing and thus in
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001754 poor performance and scalability. Use READ_ONCE() to prevent
Paul E. McKenney692118d2013-12-11 13:59:07 -08001755 invented loads.
1756
1757 (*) For aligned memory locations whose size allows them to be accessed
1758 with a single memory-reference instruction, prevents "load tearing"
1759 and "store tearing," in which a single large access is replaced by
1760 multiple smaller accesses. For example, given an architecture having
1761 16-bit store instructions with 7-bit immediate fields, the compiler
1762 might be tempted to use two 16-bit store-immediate instructions to
1763 implement the following 32-bit store:
1764
1765 p = 0x00010002;
1766
1767 Please note that GCC really does use this sort of optimization,
1768 which is not surprising given that it would likely take more
1769 than two instructions to build the constant and then store it.
1770 This optimization can therefore be a win in single-threaded code.
1771 In fact, a recent bug (since fixed) caused GCC to incorrectly use
1772 this optimization in a volatile store. In the absence of such bugs,
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001773 use of WRITE_ONCE() prevents store tearing in the following example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001774
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001775 WRITE_ONCE(p, 0x00010002);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001776
1777 Use of packed structures can also result in load and store tearing,
1778 as in this example:
1779
1780 struct __attribute__((__packed__)) foo {
1781 short a;
1782 int b;
1783 short c;
1784 };
1785 struct foo foo1, foo2;
1786 ...
1787
1788 foo2.a = foo1.a;
1789 foo2.b = foo1.b;
1790 foo2.c = foo1.c;
1791
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001792 Because there are no READ_ONCE() or WRITE_ONCE() wrappers and no
1793 volatile markings, the compiler would be well within its rights to
1794 implement these three assignment statements as a pair of 32-bit
1795 loads followed by a pair of 32-bit stores. This would result in
1796 load tearing on 'foo1.b' and store tearing on 'foo2.b'. READ_ONCE()
1797 and WRITE_ONCE() again prevent tearing in this example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001798
1799 foo2.a = foo1.a;
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001800 WRITE_ONCE(foo2.b, READ_ONCE(foo1.b));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001801 foo2.c = foo1.c;
1802
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001803All that aside, it is never necessary to use READ_ONCE() and
1804WRITE_ONCE() on a variable that has been marked volatile. For example,
1805because 'jiffies' is marked volatile, it is never necessary to
1806say READ_ONCE(jiffies). The reason for this is that READ_ONCE() and
1807WRITE_ONCE() are implemented as volatile casts, which has no effect when
1808its argument is already marked volatile.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001809
1810Please note that these compiler barriers have no direct effect on the CPU,
1811which may then reorder things however it wishes.
David Howells108b42b2006-03-31 16:00:29 +01001812
1813
1814CPU MEMORY BARRIERS
1815-------------------
1816
1817The Linux kernel has eight basic CPU memory barriers:
1818
1819 TYPE MANDATORY SMP CONDITIONAL
1820 =============== ======================= ===========================
1821 GENERAL mb() smp_mb()
1822 WRITE wmb() smp_wmb()
1823 READ rmb() smp_rmb()
1824 DATA DEPENDENCY read_barrier_depends() smp_read_barrier_depends()
1825
1826
Nick Piggin73f10282008-05-14 06:35:11 +02001827All memory barriers except the data dependency barriers imply a compiler
SeongJae Park0b6fa342016-04-12 08:52:53 -07001828barrier. Data dependencies do not impose any additional compiler ordering.
Nick Piggin73f10282008-05-14 06:35:11 +02001829
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001830Aside: In the case of data dependencies, the compiler would be expected
1831to issue the loads in the correct order (eg. `a[b]` would have to load
1832the value of b before loading a[b]), however there is no guarantee in
1833the C specification that the compiler may not speculate the value of b
1834(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
SeongJae Park0b6fa342016-04-12 08:52:53 -07001835tmp = a[b]; ). There is also the problem of a compiler reloading b after
1836having loaded a[b], thus having a newer copy of b than a[b]. A consensus
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001837has not yet been reached about these problems, however the READ_ONCE()
1838macro is a good place to start looking.
David Howells108b42b2006-03-31 16:00:29 +01001839
1840SMP memory barriers are reduced to compiler barriers on uniprocessor compiled
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001841systems because it is assumed that a CPU will appear to be self-consistent,
David Howells108b42b2006-03-31 16:00:29 +01001842and will order overlapping accesses correctly with respect to itself.
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001843However, see the subsection on "Virtual Machine Guests" below.
David Howells108b42b2006-03-31 16:00:29 +01001844
1845[!] Note that SMP memory barriers _must_ be used to control the ordering of
1846references to shared memory on SMP systems, though the use of locking instead
1847is sufficient.
1848
1849Mandatory barriers should not be used to control SMP effects, since mandatory
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001850barriers impose unnecessary overhead on both SMP and UP systems. They may,
1851however, be used to control MMIO effects on accesses through relaxed memory I/O
1852windows. These barriers are required even on non-SMP systems as they affect
1853the order in which memory operations appear to a device by prohibiting both the
1854compiler and the CPU from reordering them.
David Howells108b42b2006-03-31 16:00:29 +01001855
1856
1857There are some more advanced barrier functions:
1858
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001859 (*) smp_store_mb(var, value)
David Howells108b42b2006-03-31 16:00:29 +01001860
Oleg Nesterov75b2bd52006-11-08 17:44:38 -08001861 This assigns the value to the variable and then inserts a full memory
Davidlohr Bueso2d142e52015-10-27 12:53:51 -07001862 barrier after it. It isn't guaranteed to insert anything more than a
1863 compiler barrier in a UP compilation.
David Howells108b42b2006-03-31 16:00:29 +01001864
1865
Peter Zijlstra1b156112014-03-13 19:00:35 +01001866 (*) smp_mb__before_atomic();
1867 (*) smp_mb__after_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001868
Peter Zijlstra1b156112014-03-13 19:00:35 +01001869 These are for use with atomic (such as add, subtract, increment and
1870 decrement) functions that don't return a value, especially when used for
1871 reference counting. These functions do not imply memory barriers.
1872
1873 These are also used for atomic bitop functions that do not return a
1874 value (such as set_bit and clear_bit).
David Howells108b42b2006-03-31 16:00:29 +01001875
1876 As an example, consider a piece of code that marks an object as being dead
1877 and then decrements the object's reference count:
1878
1879 obj->dead = 1;
Peter Zijlstra1b156112014-03-13 19:00:35 +01001880 smp_mb__before_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001881 atomic_dec(&obj->ref_count);
1882
1883 This makes sure that the death mark on the object is perceived to be set
1884 *before* the reference counter is decremented.
1885
Peter Zijlstra706eeb32017-06-12 14:50:27 +02001886 See Documentation/atomic_{t,bitops}.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01001887
1888
Paul E. McKenneyad2ad5d2015-09-17 08:18:32 -07001889 (*) lockless_dereference();
SeongJae Park0b6fa342016-04-12 08:52:53 -07001890
Paul E. McKenneyad2ad5d2015-09-17 08:18:32 -07001891 This can be thought of as a pointer-fetch wrapper around the
1892 smp_read_barrier_depends() data-dependency barrier.
1893
1894 This is also similar to rcu_dereference(), but in cases where
1895 object lifetime is handled by some mechanism other than RCU, for
1896 example, when the objects removed only when the system goes down.
1897 In addition, lockless_dereference() is used in some data structures
1898 that can be used both with and without RCU.
1899
1900
Alexander Duyck1077fa32014-12-11 15:02:06 -08001901 (*) dma_wmb();
1902 (*) dma_rmb();
1903
1904 These are for use with consistent memory to guarantee the ordering
1905 of writes or reads of shared memory accessible to both the CPU and a
1906 DMA capable device.
1907
1908 For example, consider a device driver that shares memory with a device
1909 and uses a descriptor status value to indicate if the descriptor belongs
1910 to the device or the CPU, and a doorbell to notify it when new
1911 descriptors are available:
1912
1913 if (desc->status != DEVICE_OWN) {
1914 /* do not read data until we own descriptor */
1915 dma_rmb();
1916
1917 /* read/modify data */
1918 read_data = desc->data;
1919 desc->data = write_data;
1920
1921 /* flush modifications before status update */
1922 dma_wmb();
1923
1924 /* assign ownership */
1925 desc->status = DEVICE_OWN;
1926
1927 /* force memory to sync before notifying device via MMIO */
1928 wmb();
1929
1930 /* notify device of new descriptors */
1931 writel(DESC_NOTIFY, doorbell);
1932 }
1933
1934 The dma_rmb() allows us guarantee the device has released ownership
Sylvain Trias7a458002015-04-08 10:27:57 +02001935 before we read the data from the descriptor, and the dma_wmb() allows
Alexander Duyck1077fa32014-12-11 15:02:06 -08001936 us to guarantee the data is written to the descriptor before the device
1937 can see it now has ownership. The wmb() is needed to guarantee that the
1938 cache coherent memory writes have completed before attempting a write to
1939 the cache incoherent MMIO region.
1940
1941 See Documentation/DMA-API.txt for more information on consistent memory.
1942
SeongJae Parkdfeccea2016-08-11 11:17:40 -07001943
David Howells108b42b2006-03-31 16:00:29 +01001944MMIO WRITE BARRIER
1945------------------
1946
1947The Linux kernel also has a special barrier for use with memory-mapped I/O
1948writes:
1949
1950 mmiowb();
1951
1952This is a variation on the mandatory write barrier that causes writes to weakly
1953ordered I/O regions to be partially ordered. Its effects may go beyond the
1954CPU->Hardware interface and actually affect the hardware at some level.
1955
SeongJae Park166bda72016-04-12 08:52:50 -07001956See the subsection "Acquires vs I/O accesses" for more information.
David Howells108b42b2006-03-31 16:00:29 +01001957
1958
1959===============================
1960IMPLICIT KERNEL MEMORY BARRIERS
1961===============================
1962
1963Some of the other functions in the linux kernel imply memory barriers, amongst
David Howells670bd952006-06-10 09:54:12 -07001964which are locking and scheduling functions.
David Howells108b42b2006-03-31 16:00:29 +01001965
1966This specification is a _minimum_ guarantee; any particular architecture may
1967provide more substantial guarantees, but these may not be relied upon outside
1968of arch specific code.
1969
1970
SeongJae Park166bda72016-04-12 08:52:50 -07001971LOCK ACQUISITION FUNCTIONS
1972--------------------------
David Howells108b42b2006-03-31 16:00:29 +01001973
1974The Linux kernel has a number of locking constructs:
1975
1976 (*) spin locks
1977 (*) R/W spin locks
1978 (*) mutexes
1979 (*) semaphores
1980 (*) R/W semaphores
David Howells108b42b2006-03-31 16:00:29 +01001981
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001982In all cases there are variants on "ACQUIRE" operations and "RELEASE" operations
David Howells108b42b2006-03-31 16:00:29 +01001983for each construct. These operations all imply certain barriers:
1984
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001985 (1) ACQUIRE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001986
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001987 Memory operations issued after the ACQUIRE will be completed after the
1988 ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001989
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001990 Memory operations issued before the ACQUIRE may be completed after
Peter Zijlstraa9668cd2017-06-07 17:51:27 +02001991 the ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001992
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001993 (2) RELEASE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001994
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001995 Memory operations issued before the RELEASE will be completed before the
1996 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001997
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001998 Memory operations issued after the RELEASE may be completed before the
1999 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01002000
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002001 (3) ACQUIRE vs ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01002002
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002003 All ACQUIRE operations issued before another ACQUIRE operation will be
2004 completed before that ACQUIRE operation.
David Howells108b42b2006-03-31 16:00:29 +01002005
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002006 (4) ACQUIRE vs RELEASE implication:
David Howells108b42b2006-03-31 16:00:29 +01002007
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002008 All ACQUIRE operations issued before a RELEASE operation will be
2009 completed before the RELEASE operation.
David Howells108b42b2006-03-31 16:00:29 +01002010
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002011 (5) Failed conditional ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01002012
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002013 Certain locking variants of the ACQUIRE operation may fail, either due to
2014 being unable to get the lock immediately, or due to receiving an unblocked
David Howells108b42b2006-03-31 16:00:29 +01002015 signal whilst asleep waiting for the lock to become available. Failed
2016 locks do not imply any sort of barrier.
2017
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002018[!] Note: one of the consequences of lock ACQUIREs and RELEASEs being only
2019one-way barriers is that the effects of instructions outside of a critical
2020section may seep into the inside of the critical section.
David Howells108b42b2006-03-31 16:00:29 +01002021
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002022An ACQUIRE followed by a RELEASE may not be assumed to be full memory barrier
2023because it is possible for an access preceding the ACQUIRE to happen after the
2024ACQUIRE, and an access following the RELEASE to happen before the RELEASE, and
2025the two accesses can themselves then cross:
David Howells670bd952006-06-10 09:54:12 -07002026
2027 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002028 ACQUIRE M
2029 RELEASE M
David Howells670bd952006-06-10 09:54:12 -07002030 *B = b;
2031
2032may occur as:
2033
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002034 ACQUIRE M, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002035
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002036When the ACQUIRE and RELEASE are a lock acquisition and release,
2037respectively, this same reordering can occur if the lock's ACQUIRE and
2038RELEASE are to the same lock variable, but only from the perspective of
2039another CPU not holding that lock. In short, a ACQUIRE followed by an
2040RELEASE may -not- be assumed to be a full memory barrier.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002041
Paul E. McKenney12d560f2015-07-14 18:35:23 -07002042Similarly, the reverse case of a RELEASE followed by an ACQUIRE does
2043not imply a full memory barrier. Therefore, the CPU's execution of the
2044critical sections corresponding to the RELEASE and the ACQUIRE can cross,
2045so that:
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002046
2047 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002048 RELEASE M
2049 ACQUIRE N
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002050 *B = b;
2051
2052could occur as:
2053
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002054 ACQUIRE N, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002055
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002056It might appear that this reordering could introduce a deadlock.
2057However, this cannot happen because if such a deadlock threatened,
2058the RELEASE would simply complete, thereby avoiding the deadlock.
2059
2060 Why does this work?
2061
2062 One key point is that we are only talking about the CPU doing
2063 the reordering, not the compiler. If the compiler (or, for
2064 that matter, the developer) switched the operations, deadlock
2065 -could- occur.
2066
2067 But suppose the CPU reordered the operations. In this case,
2068 the unlock precedes the lock in the assembly code. The CPU
2069 simply elected to try executing the later lock operation first.
2070 If there is a deadlock, this lock operation will simply spin (or
2071 try to sleep, but more on that later). The CPU will eventually
2072 execute the unlock operation (which preceded the lock operation
2073 in the assembly code), which will unravel the potential deadlock,
2074 allowing the lock operation to succeed.
2075
2076 But what if the lock is a sleeplock? In that case, the code will
2077 try to enter the scheduler, where it will eventually encounter
2078 a memory barrier, which will force the earlier unlock operation
2079 to complete, again unraveling the deadlock. There might be
2080 a sleep-unlock race, but the locking primitive needs to resolve
2081 such races properly in any case.
2082
David Howells108b42b2006-03-31 16:00:29 +01002083Locks and semaphores may not provide any guarantee of ordering on UP compiled
2084systems, and so cannot be counted on in such a situation to actually achieve
2085anything at all - especially with respect to I/O accesses - unless combined
2086with interrupt disabling operations.
2087
SeongJae Parkd7cab362016-08-11 11:17:41 -07002088See also the section on "Inter-CPU acquiring barrier effects".
David Howells108b42b2006-03-31 16:00:29 +01002089
2090
2091As an example, consider the following:
2092
2093 *A = a;
2094 *B = b;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002095 ACQUIRE
David Howells108b42b2006-03-31 16:00:29 +01002096 *C = c;
2097 *D = d;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002098 RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002099 *E = e;
2100 *F = f;
2101
2102The following sequence of events is acceptable:
2103
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002104 ACQUIRE, {*F,*A}, *E, {*C,*D}, *B, RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002105
2106 [+] Note that {*F,*A} indicates a combined access.
2107
2108But none of the following are:
2109
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002110 {*F,*A}, *B, ACQUIRE, *C, *D, RELEASE, *E
2111 *A, *B, *C, ACQUIRE, *D, RELEASE, *E, *F
2112 *A, *B, ACQUIRE, *C, RELEASE, *D, *E, *F
2113 *B, ACQUIRE, *C, *D, RELEASE, {*F,*A}, *E
David Howells108b42b2006-03-31 16:00:29 +01002114
2115
2116
2117INTERRUPT DISABLING FUNCTIONS
2118-----------------------------
2119
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002120Functions that disable interrupts (ACQUIRE equivalent) and enable interrupts
2121(RELEASE equivalent) will act as compiler barriers only. So if memory or I/O
David Howells108b42b2006-03-31 16:00:29 +01002122barriers are required in such a situation, they must be provided from some
2123other means.
2124
2125
David Howells50fa6102009-04-28 15:01:38 +01002126SLEEP AND WAKE-UP FUNCTIONS
2127---------------------------
2128
2129Sleeping and waking on an event flagged in global data can be viewed as an
2130interaction between two pieces of data: the task state of the task waiting for
2131the event and the global data used to indicate the event. To make sure that
2132these appear to happen in the right order, the primitives to begin the process
2133of going to sleep, and the primitives to initiate a wake up imply certain
2134barriers.
2135
2136Firstly, the sleeper normally follows something like this sequence of events:
2137
2138 for (;;) {
2139 set_current_state(TASK_UNINTERRUPTIBLE);
2140 if (event_indicated)
2141 break;
2142 schedule();
2143 }
2144
2145A general memory barrier is interpolated automatically by set_current_state()
2146after it has altered the task state:
2147
2148 CPU 1
2149 ===============================
2150 set_current_state();
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002151 smp_store_mb();
David Howells50fa6102009-04-28 15:01:38 +01002152 STORE current->state
2153 <general barrier>
2154 LOAD event_indicated
2155
2156set_current_state() may be wrapped by:
2157
2158 prepare_to_wait();
2159 prepare_to_wait_exclusive();
2160
2161which therefore also imply a general memory barrier after setting the state.
2162The whole sequence above is available in various canned forms, all of which
2163interpolate the memory barrier in the right place:
2164
2165 wait_event();
2166 wait_event_interruptible();
2167 wait_event_interruptible_exclusive();
2168 wait_event_interruptible_timeout();
2169 wait_event_killable();
2170 wait_event_timeout();
2171 wait_on_bit();
2172 wait_on_bit_lock();
2173
2174
2175Secondly, code that performs a wake up normally follows something like this:
2176
2177 event_indicated = 1;
2178 wake_up(&event_wait_queue);
2179
2180or:
2181
2182 event_indicated = 1;
2183 wake_up_process(event_daemon);
2184
SeongJae Park0b6fa342016-04-12 08:52:53 -07002185A write memory barrier is implied by wake_up() and co. if and only if they
2186wake something up. The barrier occurs before the task state is cleared, and so
2187sits between the STORE to indicate the event and the STORE to set TASK_RUNNING:
David Howells50fa6102009-04-28 15:01:38 +01002188
2189 CPU 1 CPU 2
2190 =============================== ===============================
2191 set_current_state(); STORE event_indicated
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002192 smp_store_mb(); wake_up();
David Howells50fa6102009-04-28 15:01:38 +01002193 STORE current->state <write barrier>
2194 <general barrier> STORE current->state
2195 LOAD event_indicated
2196
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002197To repeat, this write memory barrier is present if and only if something
2198is actually awakened. To see this, consider the following sequence of
2199events, where X and Y are both initially zero:
2200
2201 CPU 1 CPU 2
2202 =============================== ===============================
2203 X = 1; STORE event_indicated
2204 smp_mb(); wake_up();
2205 Y = 1; wait_event(wq, Y == 1);
2206 wake_up(); load from Y sees 1, no memory barrier
2207 load from X might see 0
2208
2209In contrast, if a wakeup does occur, CPU 2's load from X would be guaranteed
2210to see 1.
2211
David Howells50fa6102009-04-28 15:01:38 +01002212The available waker functions include:
2213
2214 complete();
2215 wake_up();
2216 wake_up_all();
2217 wake_up_bit();
2218 wake_up_interruptible();
2219 wake_up_interruptible_all();
2220 wake_up_interruptible_nr();
2221 wake_up_interruptible_poll();
2222 wake_up_interruptible_sync();
2223 wake_up_interruptible_sync_poll();
2224 wake_up_locked();
2225 wake_up_locked_poll();
2226 wake_up_nr();
2227 wake_up_poll();
2228 wake_up_process();
2229
2230
2231[!] Note that the memory barriers implied by the sleeper and the waker do _not_
2232order multiple stores before the wake-up with respect to loads of those stored
2233values after the sleeper has called set_current_state(). For instance, if the
2234sleeper does:
2235
2236 set_current_state(TASK_INTERRUPTIBLE);
2237 if (event_indicated)
2238 break;
2239 __set_current_state(TASK_RUNNING);
2240 do_something(my_data);
2241
2242and the waker does:
2243
2244 my_data = value;
2245 event_indicated = 1;
2246 wake_up(&event_wait_queue);
2247
2248there's no guarantee that the change to event_indicated will be perceived by
2249the sleeper as coming after the change to my_data. In such a circumstance, the
2250code on both sides must interpolate its own memory barriers between the
2251separate data accesses. Thus the above sleeper ought to do:
2252
2253 set_current_state(TASK_INTERRUPTIBLE);
2254 if (event_indicated) {
2255 smp_rmb();
2256 do_something(my_data);
2257 }
2258
2259and the waker should do:
2260
2261 my_data = value;
2262 smp_wmb();
2263 event_indicated = 1;
2264 wake_up(&event_wait_queue);
2265
2266
David Howells108b42b2006-03-31 16:00:29 +01002267MISCELLANEOUS FUNCTIONS
2268-----------------------
2269
2270Other functions that imply barriers:
2271
2272 (*) schedule() and similar imply full memory barriers.
2273
David Howells108b42b2006-03-31 16:00:29 +01002274
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002275===================================
2276INTER-CPU ACQUIRING BARRIER EFFECTS
2277===================================
David Howells108b42b2006-03-31 16:00:29 +01002278
2279On SMP systems locking primitives give a more substantial form of barrier: one
2280that does affect memory access ordering on other CPUs, within the context of
2281conflict on any particular lock.
2282
2283
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002284ACQUIRES VS MEMORY ACCESSES
2285---------------------------
David Howells108b42b2006-03-31 16:00:29 +01002286
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002287Consider the following: the system has a pair of spinlocks (M) and (Q), and
David Howells108b42b2006-03-31 16:00:29 +01002288three CPUs; then should the following sequence of events occur:
2289
2290 CPU 1 CPU 2
2291 =============================== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002292 WRITE_ONCE(*A, a); WRITE_ONCE(*E, e);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002293 ACQUIRE M ACQUIRE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002294 WRITE_ONCE(*B, b); WRITE_ONCE(*F, f);
2295 WRITE_ONCE(*C, c); WRITE_ONCE(*G, g);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002296 RELEASE M RELEASE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002297 WRITE_ONCE(*D, d); WRITE_ONCE(*H, h);
David Howells108b42b2006-03-31 16:00:29 +01002298
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002299Then there is no guarantee as to what order CPU 3 will see the accesses to *A
David Howells108b42b2006-03-31 16:00:29 +01002300through *H occur in, other than the constraints imposed by the separate locks
SeongJae Park0b6fa342016-04-12 08:52:53 -07002301on the separate CPUs. It might, for example, see:
David Howells108b42b2006-03-31 16:00:29 +01002302
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002303 *E, ACQUIRE M, ACQUIRE Q, *G, *C, *F, *A, *B, RELEASE Q, *D, *H, RELEASE M
David Howells108b42b2006-03-31 16:00:29 +01002304
2305But it won't see any of:
2306
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002307 *B, *C or *D preceding ACQUIRE M
2308 *A, *B or *C following RELEASE M
2309 *F, *G or *H preceding ACQUIRE Q
2310 *E, *F or *G following RELEASE Q
David Howells108b42b2006-03-31 16:00:29 +01002311
2312
David Howells108b42b2006-03-31 16:00:29 +01002313
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002314ACQUIRES VS I/O ACCESSES
2315------------------------
David Howells108b42b2006-03-31 16:00:29 +01002316
2317Under certain circumstances (especially involving NUMA), I/O accesses within
2318two spinlocked sections on two different CPUs may be seen as interleaved by the
2319PCI bridge, because the PCI bridge does not necessarily participate in the
2320cache-coherence protocol, and is therefore incapable of issuing the required
2321read memory barriers.
2322
2323For example:
2324
2325 CPU 1 CPU 2
2326 =============================== ===============================
2327 spin_lock(Q)
2328 writel(0, ADDR)
2329 writel(1, DATA);
2330 spin_unlock(Q);
2331 spin_lock(Q);
2332 writel(4, ADDR);
2333 writel(5, DATA);
2334 spin_unlock(Q);
2335
2336may be seen by the PCI bridge as follows:
2337
2338 STORE *ADDR = 0, STORE *ADDR = 4, STORE *DATA = 1, STORE *DATA = 5
2339
2340which would probably cause the hardware to malfunction.
2341
2342
2343What is necessary here is to intervene with an mmiowb() before dropping the
2344spinlock, for example:
2345
2346 CPU 1 CPU 2
2347 =============================== ===============================
2348 spin_lock(Q)
2349 writel(0, ADDR)
2350 writel(1, DATA);
2351 mmiowb();
2352 spin_unlock(Q);
2353 spin_lock(Q);
2354 writel(4, ADDR);
2355 writel(5, DATA);
2356 mmiowb();
2357 spin_unlock(Q);
2358
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002359this will ensure that the two stores issued on CPU 1 appear at the PCI bridge
2360before either of the stores issued on CPU 2.
David Howells108b42b2006-03-31 16:00:29 +01002361
2362
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002363Furthermore, following a store by a load from the same device obviates the need
2364for the mmiowb(), because the load forces the store to complete before the load
David Howells108b42b2006-03-31 16:00:29 +01002365is performed:
2366
2367 CPU 1 CPU 2
2368 =============================== ===============================
2369 spin_lock(Q)
2370 writel(0, ADDR)
2371 a = readl(DATA);
2372 spin_unlock(Q);
2373 spin_lock(Q);
2374 writel(4, ADDR);
2375 b = readl(DATA);
2376 spin_unlock(Q);
2377
2378
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002379See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002380
2381
2382=================================
2383WHERE ARE MEMORY BARRIERS NEEDED?
2384=================================
2385
2386Under normal operation, memory operation reordering is generally not going to
2387be a problem as a single-threaded linear piece of code will still appear to
David Howells50fa6102009-04-28 15:01:38 +01002388work correctly, even if it's in an SMP kernel. There are, however, four
David Howells108b42b2006-03-31 16:00:29 +01002389circumstances in which reordering definitely _could_ be a problem:
2390
2391 (*) Interprocessor interaction.
2392
2393 (*) Atomic operations.
2394
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002395 (*) Accessing devices.
David Howells108b42b2006-03-31 16:00:29 +01002396
2397 (*) Interrupts.
2398
2399
2400INTERPROCESSOR INTERACTION
2401--------------------------
2402
2403When there's a system with more than one processor, more than one CPU in the
2404system may be working on the same data set at the same time. This can cause
2405synchronisation problems, and the usual way of dealing with them is to use
2406locks. Locks, however, are quite expensive, and so it may be preferable to
2407operate without the use of a lock if at all possible. In such a case
2408operations that affect both CPUs may have to be carefully ordered to prevent
2409a malfunction.
2410
2411Consider, for example, the R/W semaphore slow path. Here a waiting process is
2412queued on the semaphore, by virtue of it having a piece of its stack linked to
2413the semaphore's list of waiting processes:
2414
2415 struct rw_semaphore {
2416 ...
2417 spinlock_t lock;
2418 struct list_head waiters;
2419 };
2420
2421 struct rwsem_waiter {
2422 struct list_head list;
2423 struct task_struct *task;
2424 };
2425
2426To wake up a particular waiter, the up_read() or up_write() functions have to:
2427
2428 (1) read the next pointer from this waiter's record to know as to where the
2429 next waiter record is;
2430
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002431 (2) read the pointer to the waiter's task structure;
David Howells108b42b2006-03-31 16:00:29 +01002432
2433 (3) clear the task pointer to tell the waiter it has been given the semaphore;
2434
2435 (4) call wake_up_process() on the task; and
2436
2437 (5) release the reference held on the waiter's task struct.
2438
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002439In other words, it has to perform this sequence of events:
David Howells108b42b2006-03-31 16:00:29 +01002440
2441 LOAD waiter->list.next;
2442 LOAD waiter->task;
2443 STORE waiter->task;
2444 CALL wakeup
2445 RELEASE task
2446
2447and if any of these steps occur out of order, then the whole thing may
2448malfunction.
2449
2450Once it has queued itself and dropped the semaphore lock, the waiter does not
2451get the lock again; it instead just waits for its task pointer to be cleared
2452before proceeding. Since the record is on the waiter's stack, this means that
2453if the task pointer is cleared _before_ the next pointer in the list is read,
2454another CPU might start processing the waiter and might clobber the waiter's
2455stack before the up*() function has a chance to read the next pointer.
2456
2457Consider then what might happen to the above sequence of events:
2458
2459 CPU 1 CPU 2
2460 =============================== ===============================
2461 down_xxx()
2462 Queue waiter
2463 Sleep
2464 up_yyy()
2465 LOAD waiter->task;
2466 STORE waiter->task;
2467 Woken up by other event
2468 <preempt>
2469 Resume processing
2470 down_xxx() returns
2471 call foo()
2472 foo() clobbers *waiter
2473 </preempt>
2474 LOAD waiter->list.next;
2475 --- OOPS ---
2476
2477This could be dealt with using the semaphore lock, but then the down_xxx()
2478function has to needlessly get the spinlock again after being woken up.
2479
2480The way to deal with this is to insert a general SMP memory barrier:
2481
2482 LOAD waiter->list.next;
2483 LOAD waiter->task;
2484 smp_mb();
2485 STORE waiter->task;
2486 CALL wakeup
2487 RELEASE task
2488
2489In this case, the barrier makes a guarantee that all memory accesses before the
2490barrier will appear to happen before all the memory accesses after the barrier
2491with respect to the other CPUs on the system. It does _not_ guarantee that all
2492the memory accesses before the barrier will be complete by the time the barrier
2493instruction itself is complete.
2494
2495On a UP system - where this wouldn't be a problem - the smp_mb() is just a
2496compiler barrier, thus making sure the compiler emits the instructions in the
David Howells6bc39272006-06-25 05:49:22 -07002497right order without actually intervening in the CPU. Since there's only one
2498CPU, that CPU's dependency ordering logic will take care of everything else.
David Howells108b42b2006-03-31 16:00:29 +01002499
2500
2501ATOMIC OPERATIONS
2502-----------------
2503
David Howellsdbc87002006-04-10 22:54:23 -07002504Whilst they are technically interprocessor interaction considerations, atomic
2505operations are noted specially as some of them imply full memory barriers and
2506some don't, but they're very heavily relied on as a group throughout the
2507kernel.
2508
Peter Zijlstra706eeb32017-06-12 14:50:27 +02002509See Documentation/atomic_t.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01002510
2511
2512ACCESSING DEVICES
2513-----------------
2514
2515Many devices can be memory mapped, and so appear to the CPU as if they're just
2516a set of memory locations. To control such a device, the driver usually has to
2517make the right memory accesses in exactly the right order.
2518
2519However, having a clever CPU or a clever compiler creates a potential problem
2520in that the carefully sequenced accesses in the driver code won't reach the
2521device in the requisite order if the CPU or the compiler thinks it is more
2522efficient to reorder, combine or merge accesses - something that would cause
2523the device to malfunction.
2524
2525Inside of the Linux kernel, I/O should be done through the appropriate accessor
2526routines - such as inb() or writel() - which know how to make such accesses
2527appropriately sequential. Whilst this, for the most part, renders the explicit
2528use of memory barriers unnecessary, there are a couple of situations where they
2529might be needed:
2530
2531 (1) On some systems, I/O stores are not strongly ordered across all CPUs, and
2532 so for _all_ general drivers locks should be used and mmiowb() must be
2533 issued prior to unlocking the critical section.
2534
2535 (2) If the accessor functions are used to refer to an I/O memory window with
2536 relaxed memory access properties, then _mandatory_ memory barriers are
2537 required to enforce ordering.
2538
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002539See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002540
2541
2542INTERRUPTS
2543----------
2544
2545A driver may be interrupted by its own interrupt service routine, and thus the
2546two parts of the driver may interfere with each other's attempts to control or
2547access the device.
2548
2549This may be alleviated - at least in part - by disabling local interrupts (a
2550form of locking), such that the critical operations are all contained within
2551the interrupt-disabled section in the driver. Whilst the driver's interrupt
2552routine is executing, the driver's core may not run on the same CPU, and its
2553interrupt is not permitted to happen again until the current interrupt has been
2554handled, thus the interrupt handler does not need to lock against that.
2555
2556However, consider a driver that was talking to an ethernet card that sports an
2557address register and a data register. If that driver's core talks to the card
2558under interrupt-disablement and then the driver's interrupt handler is invoked:
2559
2560 LOCAL IRQ DISABLE
2561 writew(ADDR, 3);
2562 writew(DATA, y);
2563 LOCAL IRQ ENABLE
2564 <interrupt>
2565 writew(ADDR, 4);
2566 q = readw(DATA);
2567 </interrupt>
2568
2569The store to the data register might happen after the second store to the
2570address register if ordering rules are sufficiently relaxed:
2571
2572 STORE *ADDR = 3, STORE *ADDR = 4, STORE *DATA = y, q = LOAD *DATA
2573
2574
2575If ordering rules are relaxed, it must be assumed that accesses done inside an
2576interrupt disabled section may leak outside of it and may interleave with
2577accesses performed in an interrupt - and vice versa - unless implicit or
2578explicit barriers are used.
2579
2580Normally this won't be a problem because the I/O accesses done inside such
2581sections will include synchronous load operations on strictly ordered I/O
SeongJae Park0b6fa342016-04-12 08:52:53 -07002582registers that form implicit I/O barriers. If this isn't sufficient then an
David Howells108b42b2006-03-31 16:00:29 +01002583mmiowb() may need to be used explicitly.
2584
2585
2586A similar situation may occur between an interrupt routine and two routines
SeongJae Park0b6fa342016-04-12 08:52:53 -07002587running on separate CPUs that communicate with each other. If such a case is
David Howells108b42b2006-03-31 16:00:29 +01002588likely, then interrupt-disabling locks should be used to guarantee ordering.
2589
2590
2591==========================
2592KERNEL I/O BARRIER EFFECTS
2593==========================
2594
2595When accessing I/O memory, drivers should use the appropriate accessor
2596functions:
2597
2598 (*) inX(), outX():
2599
2600 These are intended to talk to I/O space rather than memory space, but
SeongJae Park0b6fa342016-04-12 08:52:53 -07002601 that's primarily a CPU-specific concept. The i386 and x86_64 processors
2602 do indeed have special I/O space access cycles and instructions, but many
David Howells108b42b2006-03-31 16:00:29 +01002603 CPUs don't have such a concept.
2604
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002605 The PCI bus, amongst others, defines an I/O space concept which - on such
2606 CPUs as i386 and x86_64 - readily maps to the CPU's concept of I/O
David Howells6bc39272006-06-25 05:49:22 -07002607 space. However, it may also be mapped as a virtual I/O space in the CPU's
2608 memory map, particularly on those CPUs that don't support alternate I/O
2609 spaces.
David Howells108b42b2006-03-31 16:00:29 +01002610
2611 Accesses to this space may be fully synchronous (as on i386), but
2612 intermediary bridges (such as the PCI host bridge) may not fully honour
2613 that.
2614
2615 They are guaranteed to be fully ordered with respect to each other.
2616
2617 They are not guaranteed to be fully ordered with respect to other types of
2618 memory and I/O operation.
2619
2620 (*) readX(), writeX():
2621
2622 Whether these are guaranteed to be fully ordered and uncombined with
2623 respect to each other on the issuing CPU depends on the characteristics
SeongJae Park0b6fa342016-04-12 08:52:53 -07002624 defined for the memory window through which they're accessing. On later
David Howells108b42b2006-03-31 16:00:29 +01002625 i386 architecture machines, for example, this is controlled by way of the
2626 MTRR registers.
2627
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002628 Ordinarily, these will be guaranteed to be fully ordered and uncombined,
David Howells108b42b2006-03-31 16:00:29 +01002629 provided they're not accessing a prefetchable device.
2630
2631 However, intermediary hardware (such as a PCI bridge) may indulge in
2632 deferral if it so wishes; to flush a store, a load from the same location
2633 is preferred[*], but a load from the same device or from configuration
2634 space should suffice for PCI.
2635
2636 [*] NOTE! attempting to load from the same location as was written to may
Ingo Molnare0edc782013-11-22 11:24:53 +01002637 cause a malfunction - consider the 16550 Rx/Tx serial registers for
2638 example.
David Howells108b42b2006-03-31 16:00:29 +01002639
2640 Used with prefetchable I/O memory, an mmiowb() barrier may be required to
2641 force stores to be ordered.
2642
2643 Please refer to the PCI specification for more information on interactions
2644 between PCI transactions.
2645
Will Deacona8e0aea2013-09-04 12:30:08 +01002646 (*) readX_relaxed(), writeX_relaxed()
David Howells108b42b2006-03-31 16:00:29 +01002647
Will Deacona8e0aea2013-09-04 12:30:08 +01002648 These are similar to readX() and writeX(), but provide weaker memory
SeongJae Park0b6fa342016-04-12 08:52:53 -07002649 ordering guarantees. Specifically, they do not guarantee ordering with
Will Deacona8e0aea2013-09-04 12:30:08 +01002650 respect to normal memory accesses (e.g. DMA buffers) nor do they guarantee
SeongJae Park0b6fa342016-04-12 08:52:53 -07002651 ordering with respect to LOCK or UNLOCK operations. If the latter is
2652 required, an mmiowb() barrier can be used. Note that relaxed accesses to
Will Deacona8e0aea2013-09-04 12:30:08 +01002653 the same peripheral are guaranteed to be ordered with respect to each
2654 other.
David Howells108b42b2006-03-31 16:00:29 +01002655
2656 (*) ioreadX(), iowriteX()
2657
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002658 These will perform appropriately for the type of access they're actually
David Howells108b42b2006-03-31 16:00:29 +01002659 doing, be it inX()/outX() or readX()/writeX().
2660
2661
2662========================================
2663ASSUMED MINIMUM EXECUTION ORDERING MODEL
2664========================================
2665
2666It has to be assumed that the conceptual CPU is weakly-ordered but that it will
2667maintain the appearance of program causality with respect to itself. Some CPUs
2668(such as i386 or x86_64) are more constrained than others (such as powerpc or
2669frv), and so the most relaxed case (namely DEC Alpha) must be assumed outside
2670of arch-specific code.
2671
2672This means that it must be considered that the CPU will execute its instruction
2673stream in any order it feels like - or even in parallel - provided that if an
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002674instruction in the stream depends on an earlier instruction, then that
David Howells108b42b2006-03-31 16:00:29 +01002675earlier instruction must be sufficiently complete[*] before the later
2676instruction may proceed; in other words: provided that the appearance of
2677causality is maintained.
2678
2679 [*] Some instructions have more than one effect - such as changing the
2680 condition codes, changing registers or changing memory - and different
2681 instructions may depend on different effects.
2682
2683A CPU may also discard any instruction sequence that winds up having no
2684ultimate effect. For example, if two adjacent instructions both load an
2685immediate value into the same register, the first may be discarded.
2686
2687
2688Similarly, it has to be assumed that compiler might reorder the instruction
2689stream in any way it sees fit, again provided the appearance of causality is
2690maintained.
2691
2692
2693============================
2694THE EFFECTS OF THE CPU CACHE
2695============================
2696
2697The way cached memory operations are perceived across the system is affected to
2698a certain extent by the caches that lie between CPUs and memory, and by the
2699memory coherence system that maintains the consistency of state in the system.
2700
2701As far as the way a CPU interacts with another part of the system through the
2702caches goes, the memory system has to include the CPU's caches, and memory
2703barriers for the most part act at the interface between the CPU and its cache
2704(memory barriers logically act on the dotted line in the following diagram):
2705
2706 <--- CPU ---> : <----------- Memory ----------->
2707 :
2708 +--------+ +--------+ : +--------+ +-----------+
2709 | | | | : | | | | +--------+
Ingo Molnare0edc782013-11-22 11:24:53 +01002710 | CPU | | Memory | : | CPU | | | | |
2711 | Core |--->| Access |----->| Cache |<-->| | | |
David Howells108b42b2006-03-31 16:00:29 +01002712 | | | Queue | : | | | |--->| Memory |
Ingo Molnare0edc782013-11-22 11:24:53 +01002713 | | | | : | | | | | |
2714 +--------+ +--------+ : +--------+ | | | |
David Howells108b42b2006-03-31 16:00:29 +01002715 : | Cache | +--------+
2716 : | Coherency |
2717 : | Mechanism | +--------+
2718 +--------+ +--------+ : +--------+ | | | |
2719 | | | | : | | | | | |
2720 | CPU | | Memory | : | CPU | | |--->| Device |
Ingo Molnare0edc782013-11-22 11:24:53 +01002721 | Core |--->| Access |----->| Cache |<-->| | | |
2722 | | | Queue | : | | | | | |
David Howells108b42b2006-03-31 16:00:29 +01002723 | | | | : | | | | +--------+
2724 +--------+ +--------+ : +--------+ +-----------+
2725 :
2726 :
2727
2728Although any particular load or store may not actually appear outside of the
2729CPU that issued it since it may have been satisfied within the CPU's own cache,
2730it will still appear as if the full memory access had taken place as far as the
2731other CPUs are concerned since the cache coherency mechanisms will migrate the
2732cacheline over to the accessing CPU and propagate the effects upon conflict.
2733
2734The CPU core may execute instructions in any order it deems fit, provided the
2735expected program causality appears to be maintained. Some of the instructions
2736generate load and store operations which then go into the queue of memory
2737accesses to be performed. The core may place these in the queue in any order
2738it wishes, and continue execution until it is forced to wait for an instruction
2739to complete.
2740
2741What memory barriers are concerned with is controlling the order in which
2742accesses cross from the CPU side of things to the memory side of things, and
2743the order in which the effects are perceived to happen by the other observers
2744in the system.
2745
2746[!] Memory barriers are _not_ needed within a given CPU, as CPUs always see
2747their own loads and stores as if they had happened in program order.
2748
2749[!] MMIO or other device accesses may bypass the cache system. This depends on
2750the properties of the memory window through which devices are accessed and/or
2751the use of any special device communication instructions the CPU may have.
2752
2753
2754CACHE COHERENCY
2755---------------
2756
2757Life isn't quite as simple as it may appear above, however: for while the
2758caches are expected to be coherent, there's no guarantee that that coherency
2759will be ordered. This means that whilst changes made on one CPU will
2760eventually become visible on all CPUs, there's no guarantee that they will
2761become apparent in the same order on those other CPUs.
2762
2763
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002764Consider dealing with a system that has a pair of CPUs (1 & 2), each of which
2765has a pair of parallel data caches (CPU 1 has A/B, and CPU 2 has C/D):
David Howells108b42b2006-03-31 16:00:29 +01002766
2767 :
2768 : +--------+
2769 : +---------+ | |
2770 +--------+ : +--->| Cache A |<------->| |
2771 | | : | +---------+ | |
2772 | CPU 1 |<---+ | |
2773 | | : | +---------+ | |
2774 +--------+ : +--->| Cache B |<------->| |
2775 : +---------+ | |
2776 : | Memory |
2777 : +---------+ | System |
2778 +--------+ : +--->| Cache C |<------->| |
2779 | | : | +---------+ | |
2780 | CPU 2 |<---+ | |
2781 | | : | +---------+ | |
2782 +--------+ : +--->| Cache D |<------->| |
2783 : +---------+ | |
2784 : +--------+
2785 :
2786
2787Imagine the system has the following properties:
2788
2789 (*) an odd-numbered cache line may be in cache A, cache C or it may still be
2790 resident in memory;
2791
2792 (*) an even-numbered cache line may be in cache B, cache D or it may still be
2793 resident in memory;
2794
2795 (*) whilst the CPU core is interrogating one cache, the other cache may be
2796 making use of the bus to access the rest of the system - perhaps to
2797 displace a dirty cacheline or to do a speculative load;
2798
2799 (*) each cache has a queue of operations that need to be applied to that cache
2800 to maintain coherency with the rest of the system;
2801
2802 (*) the coherency queue is not flushed by normal loads to lines already
2803 present in the cache, even though the contents of the queue may
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002804 potentially affect those loads.
David Howells108b42b2006-03-31 16:00:29 +01002805
2806Imagine, then, that two writes are made on the first CPU, with a write barrier
2807between them to guarantee that they will appear to reach that CPU's caches in
2808the requisite order:
2809
2810 CPU 1 CPU 2 COMMENT
2811 =============== =============== =======================================
2812 u == 0, v == 1 and p == &u, q == &u
2813 v = 2;
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002814 smp_wmb(); Make sure change to v is visible before
David Howells108b42b2006-03-31 16:00:29 +01002815 change to p
2816 <A:modify v=2> v is now in cache A exclusively
2817 p = &v;
2818 <B:modify p=&v> p is now in cache B exclusively
2819
2820The write memory barrier forces the other CPUs in the system to perceive that
2821the local CPU's caches have apparently been updated in the correct order. But
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002822now imagine that the second CPU wants to read those values:
David Howells108b42b2006-03-31 16:00:29 +01002823
2824 CPU 1 CPU 2 COMMENT
2825 =============== =============== =======================================
2826 ...
2827 q = p;
2828 x = *q;
2829
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002830The above pair of reads may then fail to happen in the expected order, as the
David Howells108b42b2006-03-31 16:00:29 +01002831cacheline holding p may get updated in one of the second CPU's caches whilst
2832the update to the cacheline holding v is delayed in the other of the second
2833CPU's caches by some other cache event:
2834
2835 CPU 1 CPU 2 COMMENT
2836 =============== =============== =======================================
2837 u == 0, v == 1 and p == &u, q == &u
2838 v = 2;
2839 smp_wmb();
2840 <A:modify v=2> <C:busy>
2841 <C:queue v=2>
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002842 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002843 <D:request p>
2844 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002845 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002846 x = *q;
2847 <C:read *q> Reads from v before v updated in cache
2848 <C:unbusy>
2849 <C:commit v=2>
2850
2851Basically, whilst both cachelines will be updated on CPU 2 eventually, there's
2852no guarantee that, without intervention, the order of update will be the same
2853as that committed on CPU 1.
2854
2855
2856To intervene, we need to interpolate a data dependency barrier or a read
2857barrier between the loads. This will force the cache to commit its coherency
2858queue before processing any further requests:
2859
2860 CPU 1 CPU 2 COMMENT
2861 =============== =============== =======================================
2862 u == 0, v == 1 and p == &u, q == &u
2863 v = 2;
2864 smp_wmb();
2865 <A:modify v=2> <C:busy>
2866 <C:queue v=2>
Paolo 'Blaisorblade' Giarrusso3fda9822006-10-19 23:28:19 -07002867 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002868 <D:request p>
2869 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002870 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002871 smp_read_barrier_depends()
2872 <C:unbusy>
2873 <C:commit v=2>
2874 x = *q;
2875 <C:read *q> Reads from v after v updated in cache
2876
2877
2878This sort of problem can be encountered on DEC Alpha processors as they have a
2879split cache that improves performance by making better use of the data bus.
2880Whilst most CPUs do imply a data dependency barrier on the read when a memory
2881access depends on a read, not all do, so it may not be relied on.
2882
2883Other CPUs may also have split caches, but must coordinate between the various
Matt LaPlante3f6dee92006-10-03 22:45:33 +02002884cachelets for normal memory accesses. The semantics of the Alpha removes the
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002885need for coordination in the absence of memory barriers.
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
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003050The Alpha defines the Linux kernel's memory barrier model.
David Howells108b42b2006-03-31 16:00:29 +01003051
3052See the subsection on "Cache Coherency" above.
3053
SeongJae Park0b6fa342016-04-12 08:52:53 -07003054
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003055VIRTUAL MACHINE GUESTS
SeongJae Park3dbf0912016-04-12 08:52:52 -07003056----------------------
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003057
3058Guests running within virtual machines might be affected by SMP effects even if
3059the guest itself is compiled without SMP support. This is an artifact of
3060interfacing with an SMP host while running an UP kernel. Using mandatory
3061barriers for this use-case would be possible but is often suboptimal.
3062
3063To handle this case optimally, low-level virt_mb() etc macros are available.
3064These have the same effect as smp_mb() etc when SMP is enabled, but generate
SeongJae Park0b6fa342016-04-12 08:52:53 -07003065identical code for SMP and non-SMP systems. For example, virtual machine guests
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003066should use virt_mb() rather than smp_mb() when synchronizing against a
3067(possibly SMP) host.
3068
3069These are equivalent to smp_mb() etc counterparts in all other respects,
3070in particular, they do not control MMIO effects: to control
3071MMIO effects, use mandatory barriers.
David Howells108b42b2006-03-31 16:00:29 +01003072
SeongJae Park0b6fa342016-04-12 08:52:53 -07003073
David Howells90fddab2010-03-24 09:43:00 +00003074============
3075EXAMPLE USES
3076============
3077
3078CIRCULAR BUFFERS
3079----------------
3080
3081Memory barriers can be used to implement circular buffering without the need
3082of a lock to serialise the producer with the consumer. See:
3083
3084 Documentation/circular-buffers.txt
3085
3086for details.
3087
3088
David Howells108b42b2006-03-31 16:00:29 +01003089==========
3090REFERENCES
3091==========
3092
3093Alpha AXP Architecture Reference Manual, Second Edition (Sites & Witek,
3094Digital Press)
3095 Chapter 5.2: Physical Address Space Characteristics
3096 Chapter 5.4: Caches and Write Buffers
3097 Chapter 5.5: Data Sharing
3098 Chapter 5.6: Read/Write Ordering
3099
3100AMD64 Architecture Programmer's Manual Volume 2: System Programming
3101 Chapter 7.1: Memory-Access Ordering
3102 Chapter 7.4: Buffering and Combining Memory Writes
3103
3104IA-32 Intel Architecture Software Developer's Manual, Volume 3:
3105System Programming Guide
3106 Chapter 7.1: Locked Atomic Operations
3107 Chapter 7.2: Memory Ordering
3108 Chapter 7.4: Serializing Instructions
3109
3110The SPARC Architecture Manual, Version 9
3111 Chapter 8: Memory Models
3112 Appendix D: Formal Specification of the Memory Models
3113 Appendix J: Programming with the Memory Models
3114
3115UltraSPARC Programmer Reference Manual
3116 Chapter 5: Memory Accesses and Cacheability
3117 Chapter 15: Sparc-V9 Memory Models
3118
3119UltraSPARC III Cu User's Manual
3120 Chapter 9: Memory Models
3121
3122UltraSPARC IIIi Processor User's Manual
3123 Chapter 8: Memory Models
3124
3125UltraSPARC Architecture 2005
3126 Chapter 9: Memory
3127 Appendix D: Formal Specifications of the Memory Models
3128
3129UltraSPARC T1 Supplement to the UltraSPARC Architecture 2005
3130 Chapter 8: Memory Models
3131 Appendix F: Caches and Cache Coherency
3132
3133Solaris Internals, Core Kernel Architecture, p63-68:
3134 Chapter 3.3: Hardware Considerations for Locks and
3135 Synchronization
3136
3137Unix Systems for Modern Architectures, Symmetric Multiprocessing and Caching
3138for Kernel Programmers:
3139 Chapter 13: Other Memory Models
3140
3141Intel Itanium Architecture Software Developer's Manual: Volume 1:
3142 Section 2.6: Speculation
3143 Section 4.4: Memory Access