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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>
David Howells108b42b2006-03-31 16:00:29 +01007
8Contents:
9
10 (*) Abstract memory access model.
11
12 - Device operations.
13 - Guarantees.
14
15 (*) What are memory barriers?
16
17 - Varieties of memory barrier.
18 - What may not be assumed about memory barriers?
19 - Data dependency barriers.
20 - Control dependencies.
21 - SMP barrier pairing.
22 - Examples of memory barrier sequences.
David Howells670bd952006-06-10 09:54:12 -070023 - Read memory barriers vs load speculation.
Paul E. McKenney241e6662011-02-10 16:54:50 -080024 - Transitivity
David Howells108b42b2006-03-31 16:00:29 +010025
26 (*) Explicit kernel barriers.
27
28 - Compiler barrier.
Jarek Poplawski81fc6322007-05-23 13:58:20 -070029 - CPU memory barriers.
David Howells108b42b2006-03-31 16:00:29 +010030 - MMIO write barrier.
31
32 (*) Implicit kernel memory barriers.
33
34 - Locking functions.
35 - Interrupt disabling functions.
David Howells50fa6102009-04-28 15:01:38 +010036 - Sleep and wake-up functions.
David Howells108b42b2006-03-31 16:00:29 +010037 - Miscellaneous functions.
38
39 (*) Inter-CPU locking barrier effects.
40
41 - Locks vs memory accesses.
42 - Locks vs I/O accesses.
43
44 (*) Where are memory barriers needed?
45
46 - Interprocessor interaction.
47 - Atomic operations.
48 - Accessing devices.
49 - Interrupts.
50
51 (*) Kernel I/O barrier effects.
52
53 (*) Assumed minimum execution ordering model.
54
55 (*) The effects of the cpu cache.
56
57 - Cache coherency.
58 - Cache coherency vs DMA.
59 - Cache coherency vs MMIO.
60
61 (*) The things CPUs get up to.
62
63 - And then there's the Alpha.
64
David Howells90fddab2010-03-24 09:43:00 +000065 (*) Example uses.
66
67 - Circular buffers.
68
David Howells108b42b2006-03-31 16:00:29 +010069 (*) References.
70
71
72============================
73ABSTRACT MEMORY ACCESS MODEL
74============================
75
76Consider the following abstract model of the system:
77
78 : :
79 : :
80 : :
81 +-------+ : +--------+ : +-------+
82 | | : | | : | |
83 | | : | | : | |
84 | CPU 1 |<----->| Memory |<----->| CPU 2 |
85 | | : | | : | |
86 | | : | | : | |
87 +-------+ : +--------+ : +-------+
88 ^ : ^ : ^
89 | : | : |
90 | : | : |
91 | : v : |
92 | : +--------+ : |
93 | : | | : |
94 | : | | : |
95 +---------->| Device |<----------+
96 : | | :
97 : | | :
98 : +--------+ :
99 : :
100
101Each CPU executes a program that generates memory access operations. In the
102abstract CPU, memory operation ordering is very relaxed, and a CPU may actually
103perform the memory operations in any order it likes, provided program causality
104appears to be maintained. Similarly, the compiler may also arrange the
105instructions it emits in any order it likes, provided it doesn't affect the
106apparent operation of the program.
107
108So in the above diagram, the effects of the memory operations performed by a
109CPU are perceived by the rest of the system as the operations cross the
110interface between the CPU and rest of the system (the dotted lines).
111
112
113For example, consider the following sequence of events:
114
115 CPU 1 CPU 2
116 =============== ===============
117 { A == 1; B == 2 }
Alexey Dobriyan615cc2c2014-06-06 14:36:41 -0700118 A = 3; x = B;
119 B = 4; y = A;
David Howells108b42b2006-03-31 16:00:29 +0100120
121The set of accesses as seen by the memory system in the middle can be arranged
122in 24 different combinations:
123
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400124 STORE A=3, STORE B=4, y=LOAD A->3, x=LOAD B->4
125 STORE A=3, STORE B=4, x=LOAD B->4, y=LOAD A->3
126 STORE A=3, y=LOAD A->3, STORE B=4, x=LOAD B->4
127 STORE A=3, y=LOAD A->3, x=LOAD B->2, STORE B=4
128 STORE A=3, x=LOAD B->2, STORE B=4, y=LOAD A->3
129 STORE A=3, x=LOAD B->2, y=LOAD A->3, STORE B=4
130 STORE B=4, STORE A=3, y=LOAD A->3, x=LOAD B->4
David Howells108b42b2006-03-31 16:00:29 +0100131 STORE B=4, ...
132 ...
133
134and can thus result in four different combinations of values:
135
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400136 x == 2, y == 1
137 x == 2, y == 3
138 x == 4, y == 1
139 x == 4, y == 3
David Howells108b42b2006-03-31 16:00:29 +0100140
141
142Furthermore, the stores committed by a CPU to the memory system may not be
143perceived by the loads made by another CPU in the same order as the stores were
144committed.
145
146
147As a further example, consider this sequence of events:
148
149 CPU 1 CPU 2
150 =============== ===============
151 { A == 1, B == 2, C = 3, P == &A, Q == &C }
152 B = 4; Q = P;
153 P = &B D = *Q;
154
155There is an obvious data dependency here, as the value loaded into D depends on
156the address retrieved from P by CPU 2. At the end of the sequence, any of the
157following results are possible:
158
159 (Q == &A) and (D == 1)
160 (Q == &B) and (D == 2)
161 (Q == &B) and (D == 4)
162
163Note that CPU 2 will never try and load C into D because the CPU will load P
164into Q before issuing the load of *Q.
165
166
167DEVICE OPERATIONS
168-----------------
169
170Some devices present their control interfaces as collections of memory
171locations, but the order in which the control registers are accessed is very
172important. For instance, imagine an ethernet card with a set of internal
173registers that are accessed through an address port register (A) and a data
174port register (D). To read internal register 5, the following code might then
175be used:
176
177 *A = 5;
178 x = *D;
179
180but this might show up as either of the following two sequences:
181
182 STORE *A = 5, x = LOAD *D
183 x = LOAD *D, STORE *A = 5
184
185the second of which will almost certainly result in a malfunction, since it set
186the address _after_ attempting to read the register.
187
188
189GUARANTEES
190----------
191
192There are some minimal guarantees that may be expected of a CPU:
193
194 (*) On any given CPU, dependent memory accesses will be issued in order, with
195 respect to itself. This means that for:
196
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700197 WRITE_ONCE(Q, P); smp_read_barrier_depends(); D = READ_ONCE(*Q);
David Howells108b42b2006-03-31 16:00:29 +0100198
199 the CPU will issue the following memory operations:
200
201 Q = LOAD P, D = LOAD *Q
202
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800203 and always in that order. On most systems, smp_read_barrier_depends()
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700204 does nothing, but it is required for DEC Alpha. The READ_ONCE()
205 and WRITE_ONCE() are required to prevent compiler mischief. Please
206 note that you should normally use something like rcu_dereference()
207 instead of open-coding smp_read_barrier_depends().
David Howells108b42b2006-03-31 16:00:29 +0100208
209 (*) Overlapping loads and stores within a particular CPU will appear to be
210 ordered within that CPU. This means that for:
211
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700212 a = READ_ONCE(*X); WRITE_ONCE(*X, b);
David Howells108b42b2006-03-31 16:00:29 +0100213
214 the CPU will only issue the following sequence of memory operations:
215
216 a = LOAD *X, STORE *X = b
217
218 And for:
219
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700220 WRITE_ONCE(*X, c); d = READ_ONCE(*X);
David Howells108b42b2006-03-31 16:00:29 +0100221
222 the CPU will only issue:
223
224 STORE *X = c, d = LOAD *X
225
Matt LaPlantefa00e7e2006-11-30 04:55:36 +0100226 (Loads and stores overlap if they are targeted at overlapping pieces of
David Howells108b42b2006-03-31 16:00:29 +0100227 memory).
228
229And there are a number of things that _must_ or _must_not_ be assumed:
230
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700231 (*) It _must_not_ be assumed that the compiler will do what you want
232 with memory references that are not protected by READ_ONCE() and
233 WRITE_ONCE(). Without them, the compiler is within its rights to
234 do all sorts of "creative" transformations, which are covered in
235 the Compiler Barrier section.
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800236
David Howells108b42b2006-03-31 16:00:29 +0100237 (*) It _must_not_ be assumed that independent loads and stores will be issued
238 in the order given. This means that for:
239
240 X = *A; Y = *B; *D = Z;
241
242 we may get any of the following sequences:
243
244 X = LOAD *A, Y = LOAD *B, STORE *D = Z
245 X = LOAD *A, STORE *D = Z, Y = LOAD *B
246 Y = LOAD *B, X = LOAD *A, STORE *D = Z
247 Y = LOAD *B, STORE *D = Z, X = LOAD *A
248 STORE *D = Z, X = LOAD *A, Y = LOAD *B
249 STORE *D = Z, Y = LOAD *B, X = LOAD *A
250
251 (*) It _must_ be assumed that overlapping memory accesses may be merged or
252 discarded. This means that for:
253
254 X = *A; Y = *(A + 4);
255
256 we may get any one of the following sequences:
257
258 X = LOAD *A; Y = LOAD *(A + 4);
259 Y = LOAD *(A + 4); X = LOAD *A;
260 {X, Y} = LOAD {*A, *(A + 4) };
261
262 And for:
263
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700264 *A = X; *(A + 4) = Y;
David Howells108b42b2006-03-31 16:00:29 +0100265
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700266 we may get any of:
David Howells108b42b2006-03-31 16:00:29 +0100267
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700268 STORE *A = X; STORE *(A + 4) = Y;
269 STORE *(A + 4) = Y; STORE *A = X;
270 STORE {*A, *(A + 4) } = {X, Y};
David Howells108b42b2006-03-31 16:00:29 +0100271
Paul E. McKenney432fbf32014-09-04 17:12:49 -0700272And there are anti-guarantees:
273
274 (*) These guarantees do not apply to bitfields, because compilers often
275 generate code to modify these using non-atomic read-modify-write
276 sequences. Do not attempt to use bitfields to synchronize parallel
277 algorithms.
278
279 (*) Even in cases where bitfields are protected by locks, all fields
280 in a given bitfield must be protected by one lock. If two fields
281 in a given bitfield are protected by different locks, the compiler's
282 non-atomic read-modify-write sequences can cause an update to one
283 field to corrupt the value of an adjacent field.
284
285 (*) These guarantees apply only to properly aligned and sized scalar
286 variables. "Properly sized" currently means variables that are
287 the same size as "char", "short", "int" and "long". "Properly
288 aligned" means the natural alignment, thus no constraints for
289 "char", two-byte alignment for "short", four-byte alignment for
290 "int", and either four-byte or eight-byte alignment for "long",
291 on 32-bit and 64-bit systems, respectively. Note that these
292 guarantees were introduced into the C11 standard, so beware when
293 using older pre-C11 compilers (for example, gcc 4.6). The portion
294 of the standard containing this guarantee is Section 3.14, which
295 defines "memory location" as follows:
296
297 memory location
298 either an object of scalar type, or a maximal sequence
299 of adjacent bit-fields all having nonzero width
300
301 NOTE 1: Two threads of execution can update and access
302 separate memory locations without interfering with
303 each other.
304
305 NOTE 2: A bit-field and an adjacent non-bit-field member
306 are in separate memory locations. The same applies
307 to two bit-fields, if one is declared inside a nested
308 structure declaration and the other is not, or if the two
309 are separated by a zero-length bit-field declaration,
310 or if they are separated by a non-bit-field member
311 declaration. It is not safe to concurrently update two
312 bit-fields in the same structure if all members declared
313 between them are also bit-fields, no matter what the
314 sizes of those intervening bit-fields happen to be.
315
David Howells108b42b2006-03-31 16:00:29 +0100316
317=========================
318WHAT ARE MEMORY BARRIERS?
319=========================
320
321As can be seen above, independent memory operations are effectively performed
322in random order, but this can be a problem for CPU-CPU interaction and for I/O.
323What is required is some way of intervening to instruct the compiler and the
324CPU to restrict the order.
325
326Memory barriers are such interventions. They impose a perceived partial
David Howells2b948952006-06-25 05:48:49 -0700327ordering over the memory operations on either side of the barrier.
328
329Such enforcement is important because the CPUs and other devices in a system
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700330can use a variety of tricks to improve performance, including reordering,
David Howells2b948952006-06-25 05:48:49 -0700331deferral and combination of memory operations; speculative loads; speculative
332branch prediction and various types of caching. Memory barriers are used to
333override or suppress these tricks, allowing the code to sanely control the
334interaction of multiple CPUs and/or devices.
David Howells108b42b2006-03-31 16:00:29 +0100335
336
337VARIETIES OF MEMORY BARRIER
338---------------------------
339
340Memory barriers come in four basic varieties:
341
342 (1) Write (or store) memory barriers.
343
344 A write memory barrier gives a guarantee that all the STORE operations
345 specified before the barrier will appear to happen before all the STORE
346 operations specified after the barrier with respect to the other
347 components of the system.
348
349 A write barrier is a partial ordering on stores only; it is not required
350 to have any effect on loads.
351
David Howells6bc39272006-06-25 05:49:22 -0700352 A CPU can be viewed as committing a sequence of store operations to the
David Howells108b42b2006-03-31 16:00:29 +0100353 memory system as time progresses. All stores before a write barrier will
354 occur in the sequence _before_ all the stores after the write barrier.
355
356 [!] Note that write barriers should normally be paired with read or data
357 dependency barriers; see the "SMP barrier pairing" subsection.
358
359
360 (2) Data dependency barriers.
361
362 A data dependency barrier is a weaker form of read barrier. In the case
363 where two loads are performed such that the second depends on the result
364 of the first (eg: the first load retrieves the address to which the second
365 load will be directed), a data dependency barrier would be required to
366 make sure that the target of the second load is updated before the address
367 obtained by the first load is accessed.
368
369 A data dependency barrier is a partial ordering on interdependent loads
370 only; it is not required to have any effect on stores, independent loads
371 or overlapping loads.
372
373 As mentioned in (1), the other CPUs in the system can be viewed as
374 committing sequences of stores to the memory system that the CPU being
375 considered can then perceive. A data dependency barrier issued by the CPU
376 under consideration guarantees that for any load preceding it, if that
377 load touches one of a sequence of stores from another CPU, then by the
378 time the barrier completes, the effects of all the stores prior to that
379 touched by the load will be perceptible to any loads issued after the data
380 dependency barrier.
381
382 See the "Examples of memory barrier sequences" subsection for diagrams
383 showing the ordering constraints.
384
385 [!] Note that the first load really has to have a _data_ dependency and
386 not a control dependency. If the address for the second load is dependent
387 on the first load, but the dependency is through a conditional rather than
388 actually loading the address itself, then it's a _control_ dependency and
389 a full read barrier or better is required. See the "Control dependencies"
390 subsection for more information.
391
392 [!] Note that data dependency barriers should normally be paired with
393 write barriers; see the "SMP barrier pairing" subsection.
394
395
396 (3) Read (or load) memory barriers.
397
398 A read barrier is a data dependency barrier plus a guarantee that all the
399 LOAD operations specified before the barrier will appear to happen before
400 all the LOAD operations specified after the barrier with respect to the
401 other components of the system.
402
403 A read barrier is a partial ordering on loads only; it is not required to
404 have any effect on stores.
405
406 Read memory barriers imply data dependency barriers, and so can substitute
407 for them.
408
409 [!] Note that read barriers should normally be paired with write barriers;
410 see the "SMP barrier pairing" subsection.
411
412
413 (4) General memory barriers.
414
David Howells670bd952006-06-10 09:54:12 -0700415 A general memory barrier gives a guarantee that all the LOAD and STORE
416 operations specified before the barrier will appear to happen before all
417 the LOAD and STORE operations specified after the barrier with respect to
418 the other components of the system.
419
420 A general memory barrier is a partial ordering over both loads and stores.
David Howells108b42b2006-03-31 16:00:29 +0100421
422 General memory barriers imply both read and write memory barriers, and so
423 can substitute for either.
424
425
426And a couple of implicit varieties:
427
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100428 (5) ACQUIRE operations.
David Howells108b42b2006-03-31 16:00:29 +0100429
430 This acts as a one-way permeable barrier. It guarantees that all memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100431 operations after the ACQUIRE operation will appear to happen after the
432 ACQUIRE operation with respect to the other components of the system.
433 ACQUIRE operations include LOCK operations and smp_load_acquire()
434 operations.
David Howells108b42b2006-03-31 16:00:29 +0100435
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100436 Memory operations that occur before an ACQUIRE operation may appear to
437 happen after it completes.
David Howells108b42b2006-03-31 16:00:29 +0100438
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100439 An ACQUIRE operation should almost always be paired with a RELEASE
440 operation.
David Howells108b42b2006-03-31 16:00:29 +0100441
442
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100443 (6) RELEASE operations.
David Howells108b42b2006-03-31 16:00:29 +0100444
445 This also acts as a one-way permeable barrier. It guarantees that all
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100446 memory operations before the RELEASE operation will appear to happen
447 before the RELEASE operation with respect to the other components of the
448 system. RELEASE operations include UNLOCK operations and
449 smp_store_release() operations.
David Howells108b42b2006-03-31 16:00:29 +0100450
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100451 Memory operations that occur after a RELEASE operation may appear to
David Howells108b42b2006-03-31 16:00:29 +0100452 happen before it completes.
453
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100454 The use of ACQUIRE and RELEASE operations generally precludes the need
455 for other sorts of memory barrier (but note the exceptions mentioned in
456 the subsection "MMIO write barrier"). In addition, a RELEASE+ACQUIRE
457 pair is -not- guaranteed to act as a full memory barrier. However, after
458 an ACQUIRE on a given variable, all memory accesses preceding any prior
459 RELEASE on that same variable are guaranteed to be visible. In other
460 words, within a given variable's critical section, all accesses of all
461 previous critical sections for that variable are guaranteed to have
462 completed.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -0800463
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100464 This means that ACQUIRE acts as a minimal "acquire" operation and
465 RELEASE acts as a minimal "release" operation.
David Howells108b42b2006-03-31 16:00:29 +0100466
467
468Memory barriers are only required where there's a possibility of interaction
469between two CPUs or between a CPU and a device. If it can be guaranteed that
470there won't be any such interaction in any particular piece of code, then
471memory barriers are unnecessary in that piece of code.
472
473
474Note that these are the _minimum_ guarantees. Different architectures may give
475more substantial guarantees, but they may _not_ be relied upon outside of arch
476specific code.
477
478
479WHAT MAY NOT BE ASSUMED ABOUT MEMORY BARRIERS?
480----------------------------------------------
481
482There are certain things that the Linux kernel memory barriers do not guarantee:
483
484 (*) There is no guarantee that any of the memory accesses specified before a
485 memory barrier will be _complete_ by the completion of a memory barrier
486 instruction; the barrier can be considered to draw a line in that CPU's
487 access queue that accesses of the appropriate type may not cross.
488
489 (*) There is no guarantee that issuing a memory barrier on one CPU will have
490 any direct effect on another CPU or any other hardware in the system. The
491 indirect effect will be the order in which the second CPU sees the effects
492 of the first CPU's accesses occur, but see the next point:
493
David Howells6bc39272006-06-25 05:49:22 -0700494 (*) There is no guarantee that a CPU will see the correct order of effects
David Howells108b42b2006-03-31 16:00:29 +0100495 from a second CPU's accesses, even _if_ the second CPU uses a memory
496 barrier, unless the first CPU _also_ uses a matching memory barrier (see
497 the subsection on "SMP Barrier Pairing").
498
499 (*) There is no guarantee that some intervening piece of off-the-CPU
500 hardware[*] will not reorder the memory accesses. CPU cache coherency
501 mechanisms should propagate the indirect effects of a memory barrier
502 between CPUs, but might not do so in order.
503
504 [*] For information on bus mastering DMA and coherency please read:
505
Randy Dunlap4b5ff462008-03-10 17:16:32 -0700506 Documentation/PCI/pci.txt
Paul Bolle395cf962011-08-15 02:02:26 +0200507 Documentation/DMA-API-HOWTO.txt
David Howells108b42b2006-03-31 16:00:29 +0100508 Documentation/DMA-API.txt
509
510
511DATA DEPENDENCY BARRIERS
512------------------------
513
514The usage requirements of data dependency barriers are a little subtle, and
515it's not always obvious that they're needed. To illustrate, consider the
516following sequence of events:
517
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800518 CPU 1 CPU 2
519 =============== ===============
David Howells108b42b2006-03-31 16:00:29 +0100520 { A == 1, B == 2, C = 3, P == &A, Q == &C }
521 B = 4;
522 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700523 WRITE_ONCE(P, &B)
524 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800525 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100526
527There's a clear data dependency here, and it would seem that by the end of the
528sequence, Q must be either &A or &B, and that:
529
530 (Q == &A) implies (D == 1)
531 (Q == &B) implies (D == 4)
532
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700533But! CPU 2's perception of P may be updated _before_ its perception of B, thus
David Howells108b42b2006-03-31 16:00:29 +0100534leading to the following situation:
535
536 (Q == &B) and (D == 2) ????
537
538Whilst this may seem like a failure of coherency or causality maintenance, it
539isn't, and this behaviour can be observed on certain real CPUs (such as the DEC
540Alpha).
541
David Howells2b948952006-06-25 05:48:49 -0700542To deal with this, a data dependency barrier or better must be inserted
543between the address load and the data load:
David Howells108b42b2006-03-31 16:00:29 +0100544
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800545 CPU 1 CPU 2
546 =============== ===============
David Howells108b42b2006-03-31 16:00:29 +0100547 { A == 1, B == 2, C = 3, P == &A, Q == &C }
548 B = 4;
549 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700550 WRITE_ONCE(P, &B);
551 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800552 <data dependency barrier>
553 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100554
555This enforces the occurrence of one of the two implications, and prevents the
556third possibility from arising.
557
558[!] Note that this extremely counterintuitive situation arises most easily on
559machines with split caches, so that, for example, one cache bank processes
560even-numbered cache lines and the other bank processes odd-numbered cache
561lines. The pointer P might be stored in an odd-numbered cache line, and the
562variable B might be stored in an even-numbered cache line. Then, if the
563even-numbered bank of the reading CPU's cache is extremely busy while the
564odd-numbered bank is idle, one can see the new value of the pointer P (&B),
David Howells6bc39272006-06-25 05:49:22 -0700565but the old value of the variable B (2).
David Howells108b42b2006-03-31 16:00:29 +0100566
567
Ingo Molnare0edc782013-11-22 11:24:53 +0100568Another example of where data dependency barriers might be required is where a
David Howells108b42b2006-03-31 16:00:29 +0100569number is read from memory and then used to calculate the index for an array
570access:
571
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800572 CPU 1 CPU 2
573 =============== ===============
David Howells108b42b2006-03-31 16:00:29 +0100574 { M[0] == 1, M[1] == 2, M[3] = 3, P == 0, Q == 3 }
575 M[1] = 4;
576 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700577 WRITE_ONCE(P, 1);
578 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800579 <data dependency barrier>
580 D = M[Q];
David Howells108b42b2006-03-31 16:00:29 +0100581
582
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800583The data dependency barrier is very important to the RCU system,
584for example. See rcu_assign_pointer() and rcu_dereference() in
585include/linux/rcupdate.h. This permits the current target of an RCU'd
586pointer to be replaced with a new modified target, without the replacement
587target appearing to be incompletely initialised.
David Howells108b42b2006-03-31 16:00:29 +0100588
589See also the subsection on "Cache Coherency" for a more thorough example.
590
591
592CONTROL DEPENDENCIES
593--------------------
594
Paul E. McKenneyff382812015-02-17 10:00:06 -0800595A load-load control dependency requires a full read memory barrier, not
596simply a data dependency barrier to make it work correctly. Consider the
597following bit of code:
David Howells108b42b2006-03-31 16:00:29 +0100598
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700599 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800600 if (q) {
601 <data dependency barrier> /* BUG: No data dependency!!! */
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700602 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700603 }
David Howells108b42b2006-03-31 16:00:29 +0100604
605This will not have the desired effect because there is no actual data
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800606dependency, but rather a control dependency that the CPU may short-circuit
607by attempting to predict the outcome in advance, so that other CPUs see
608the load from b as having happened before the load from a. In such a
609case what's actually required is:
David Howells108b42b2006-03-31 16:00:29 +0100610
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700611 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800612 if (q) {
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700613 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700614 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700615 }
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800616
617However, stores are not speculated. This means that ordering -is- provided
Paul E. McKenneyff382812015-02-17 10:00:06 -0800618for load-store control dependencies, as in the following example:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800619
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800620 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700621 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700622 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800623 }
624
Paul E. McKenney5af46922015-04-25 12:48:29 -0700625Control dependencies pair normally with other types of barriers. That
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800626said, please note that READ_ONCE() is not optional! Without the
627READ_ONCE(), the compiler might combine the load from 'a' with other
628loads from 'a', and the store to 'b' with other stores to 'b', with
629possible highly counterintuitive effects on ordering.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800630
631Worse yet, if the compiler is able to prove (say) that the value of
632variable 'a' is always non-zero, it would be well within its rights
633to optimize the original example by eliminating the "if" statement
634as follows:
635
636 q = a;
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700637 b = p; /* BUG: Compiler and CPU can both reorder!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800638
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800639So don't leave out the READ_ONCE().
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700640
641It is tempting to try to enforce ordering on identical stores on both
642branches of the "if" statement as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800643
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800644 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800645 if (q) {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800646 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700647 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800648 do_something();
649 } else {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800650 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700651 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800652 do_something_else();
653 }
654
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700655Unfortunately, current compilers will transform this as follows at high
656optimization levels:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800657
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800658 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700659 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700660 WRITE_ONCE(b, p); /* BUG: No ordering vs. load from a!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800661 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700662 /* WRITE_ONCE(b, p); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800663 do_something();
664 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700665 /* WRITE_ONCE(b, p); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800666 do_something_else();
667 }
668
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700669Now there is no conditional between the load from 'a' and the store to
670'b', which means that the CPU is within its rights to reorder them:
671The conditional is absolutely required, and must be present in the
672assembly code even after all compiler optimizations have been applied.
673Therefore, if you need ordering in this example, you need explicit
674memory barriers, for example, smp_store_release():
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800675
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700676 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700677 if (q) {
678 smp_store_release(&b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800679 do_something();
680 } else {
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700681 smp_store_release(&b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800682 do_something_else();
683 }
684
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700685In contrast, without explicit memory barriers, two-legged-if control
686ordering is guaranteed only when the stores differ, for example:
687
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800688 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700689 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700690 WRITE_ONCE(b, p);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700691 do_something();
692 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700693 WRITE_ONCE(b, r);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700694 do_something_else();
695 }
696
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800697The initial READ_ONCE() is still required to prevent the compiler from
698proving the value of 'a'.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800699
700In addition, you need to be careful what you do with the local variable 'q',
701otherwise the compiler might be able to guess the value and again remove
702the needed conditional. For example:
703
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800704 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800705 if (q % MAX) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700706 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800707 do_something();
708 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700709 WRITE_ONCE(b, r);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800710 do_something_else();
711 }
712
713If MAX is defined to be 1, then the compiler knows that (q % MAX) is
714equal to zero, in which case the compiler is within its rights to
715transform the above code into the following:
716
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800717 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700718 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800719 do_something_else();
720
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700721Given this transformation, the CPU is not required to respect the ordering
722between the load from variable 'a' and the store to variable 'b'. It is
723tempting to add a barrier(), but this does not help. The conditional
724is gone, and the barrier won't bring it back. Therefore, if you are
725relying on this ordering, you should make sure that MAX is greater than
726one, perhaps as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800727
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800728 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800729 BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
730 if (q % MAX) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700731 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800732 do_something();
733 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700734 WRITE_ONCE(b, r);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800735 do_something_else();
736 }
737
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700738Please note once again that the stores to 'b' differ. If they were
739identical, as noted earlier, the compiler could pull this store outside
740of the 'if' statement.
741
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700742You must also be careful not to rely too much on boolean short-circuit
743evaluation. Consider this example:
744
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800745 q = READ_ONCE(a);
Paul E. McKenney57aecae2015-05-18 18:27:42 -0700746 if (q || 1 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700747 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700748
Paul E. McKenney5af46922015-04-25 12:48:29 -0700749Because the first condition cannot fault and the second condition is
750always true, the compiler can transform this example as following,
751defeating control dependency:
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700752
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800753 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700754 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700755
756This example underscores the need to ensure that the compiler cannot
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700757out-guess your code. More generally, although READ_ONCE() does force
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700758the compiler to actually emit code for a given load, it does not force
759the compiler to use the results.
760
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800761Finally, control dependencies do -not- provide transitivity. This is
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700762demonstrated by two related examples, with the initial values of
763x and y both being zero:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800764
765 CPU 0 CPU 1
Paul E. McKenney5af46922015-04-25 12:48:29 -0700766 ======================= =======================
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800767 r1 = READ_ONCE(x); r2 = READ_ONCE(y);
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700768 if (r1 > 0) if (r2 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700769 WRITE_ONCE(y, 1); WRITE_ONCE(x, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800770
771 assert(!(r1 == 1 && r2 == 1));
772
773The above two-CPU example will never trigger the assert(). However,
774if control dependencies guaranteed transitivity (which they do not),
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700775then adding the following CPU would guarantee a related assertion:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800776
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700777 CPU 2
778 =====================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700779 WRITE_ONCE(x, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800780
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700781 assert(!(r1 == 2 && r2 == 1 && x == 2)); /* FAILS!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800782
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700783But because control dependencies do -not- provide transitivity, the above
784assertion can fail after the combined three-CPU example completes. If you
785need the three-CPU example to provide ordering, you will need smp_mb()
786between the loads and stores in the CPU 0 and CPU 1 code fragments,
Paul E. McKenney5af46922015-04-25 12:48:29 -0700787that is, just before or just after the "if" statements. Furthermore,
788the original two-CPU example is very fragile and should be avoided.
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700789
790These two examples are the LB and WWC litmus tests from this paper:
791http://www.cl.cam.ac.uk/users/pes20/ppc-supplemental/test6.pdf and this
792site: https://www.cl.cam.ac.uk/~pes20/ppcmem/index.html.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800793
794In summary:
795
796 (*) Control dependencies can order prior loads against later stores.
797 However, they do -not- guarantee any other sort of ordering:
798 Not prior loads against later loads, nor prior stores against
799 later anything. If you need these other forms of ordering,
Davidlohr Buesod87510c2014-12-28 01:11:16 -0800800 use smp_rmb(), smp_wmb(), or, in the case of prior stores and
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800801 later loads, smp_mb().
802
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800803 (*) If both legs of the "if" statement begin with identical stores
804 to the same variable, a barrier() statement is required at the
805 beginning of each leg of the "if" statement.
806
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800807 (*) Control dependencies require at least one run-time conditional
Paul E. McKenney586dd562014-02-11 12:28:06 -0800808 between the prior load and the subsequent store, and this
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700809 conditional must involve the prior load. If the compiler is able
810 to optimize the conditional away, it will have also optimized
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800811 away the ordering. Careful use of READ_ONCE() and WRITE_ONCE()
812 can help to preserve the needed conditional.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800813
814 (*) Control dependencies require that the compiler avoid reordering the
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800815 dependency into nonexistence. Careful use of READ_ONCE() or
816 atomic{,64}_read() can help to preserve your control dependency.
817 Please see the Compiler Barrier section for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800818
Paul E. McKenneyff382812015-02-17 10:00:06 -0800819 (*) Control dependencies pair normally with other types of barriers.
820
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800821 (*) Control dependencies do -not- provide transitivity. If you
822 need transitivity, use smp_mb().
David Howells108b42b2006-03-31 16:00:29 +0100823
824
825SMP BARRIER PAIRING
826-------------------
827
828When dealing with CPU-CPU interactions, certain types of memory barrier should
829always be paired. A lack of appropriate pairing is almost certainly an error.
830
Paul E. McKenneyff382812015-02-17 10:00:06 -0800831General barriers pair with each other, though they also pair with most
832other types of barriers, albeit without transitivity. An acquire barrier
833pairs with a release barrier, but both may also pair with other barriers,
834including of course general barriers. A write barrier pairs with a data
835dependency barrier, a control dependency, an acquire barrier, a release
836barrier, a read barrier, or a general barrier. Similarly a read barrier,
837control dependency, or a data dependency barrier pairs with a write
838barrier, an acquire barrier, a release barrier, or a general barrier:
David Howells108b42b2006-03-31 16:00:29 +0100839
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800840 CPU 1 CPU 2
841 =============== ===============
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700842 WRITE_ONCE(a, 1);
David Howells108b42b2006-03-31 16:00:29 +0100843 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700844 WRITE_ONCE(b, 2); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800845 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700846 y = READ_ONCE(a);
David Howells108b42b2006-03-31 16:00:29 +0100847
848Or:
849
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800850 CPU 1 CPU 2
851 =============== ===============================
David Howells108b42b2006-03-31 16:00:29 +0100852 a = 1;
853 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700854 WRITE_ONCE(b, &a); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800855 <data dependency barrier>
856 y = *x;
David Howells108b42b2006-03-31 16:00:29 +0100857
Paul E. McKenneyff382812015-02-17 10:00:06 -0800858Or even:
859
860 CPU 1 CPU 2
861 =============== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700862 r1 = READ_ONCE(y);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800863 <general barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700864 WRITE_ONCE(y, 1); if (r2 = READ_ONCE(x)) {
Paul E. McKenneyff382812015-02-17 10:00:06 -0800865 <implicit control dependency>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700866 WRITE_ONCE(y, 1);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800867 }
868
869 assert(r1 == 0 || r2 == 0);
870
David Howells108b42b2006-03-31 16:00:29 +0100871Basically, the read barrier always has to be there, even though it can be of
872the "weaker" type.
873
David Howells670bd952006-06-10 09:54:12 -0700874[!] Note that the stores before the write barrier would normally be expected to
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700875match the loads after the read barrier or the data dependency barrier, and vice
David Howells670bd952006-06-10 09:54:12 -0700876versa:
877
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800878 CPU 1 CPU 2
879 =================== ===================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700880 WRITE_ONCE(a, 1); }---- --->{ v = READ_ONCE(c);
881 WRITE_ONCE(b, 2); } \ / { w = READ_ONCE(d);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800882 <write barrier> \ <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700883 WRITE_ONCE(c, 3); } / \ { x = READ_ONCE(a);
884 WRITE_ONCE(d, 4); }---- --->{ y = READ_ONCE(b);
David Howells670bd952006-06-10 09:54:12 -0700885
David Howells108b42b2006-03-31 16:00:29 +0100886
887EXAMPLES OF MEMORY BARRIER SEQUENCES
888------------------------------------
889
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700890Firstly, write barriers act as partial orderings on store operations.
David Howells108b42b2006-03-31 16:00:29 +0100891Consider the following sequence of events:
892
893 CPU 1
894 =======================
895 STORE A = 1
896 STORE B = 2
897 STORE C = 3
898 <write barrier>
899 STORE D = 4
900 STORE E = 5
901
902This sequence of events is committed to the memory coherence system in an order
903that the rest of the system might perceive as the unordered set of { STORE A,
Adrian Bunk80f72282006-06-30 18:27:16 +0200904STORE B, STORE C } all occurring before the unordered set of { STORE D, STORE E
David Howells108b42b2006-03-31 16:00:29 +0100905}:
906
907 +-------+ : :
908 | | +------+
909 | |------>| C=3 | } /\
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700910 | | : +------+ }----- \ -----> Events perceptible to
911 | | : | A=1 | } \/ the rest of the system
David Howells108b42b2006-03-31 16:00:29 +0100912 | | : +------+ }
913 | CPU 1 | : | B=2 | }
914 | | +------+ }
915 | | wwwwwwwwwwwwwwww } <--- At this point the write barrier
916 | | +------+ } requires all stores prior to the
917 | | : | E=5 | } barrier to be committed before
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700918 | | : +------+ } further stores may take place
David Howells108b42b2006-03-31 16:00:29 +0100919 | |------>| D=4 | }
920 | | +------+
921 +-------+ : :
922 |
David Howells670bd952006-06-10 09:54:12 -0700923 | Sequence in which stores are committed to the
924 | memory system by CPU 1
David Howells108b42b2006-03-31 16:00:29 +0100925 V
926
927
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700928Secondly, data dependency barriers act as partial orderings on data-dependent
David Howells108b42b2006-03-31 16:00:29 +0100929loads. Consider the following sequence of events:
930
931 CPU 1 CPU 2
932 ======================= =======================
David Howellsc14038c2006-04-10 22:54:24 -0700933 { B = 7; X = 9; Y = 8; C = &Y }
David Howells108b42b2006-03-31 16:00:29 +0100934 STORE A = 1
935 STORE B = 2
936 <write barrier>
937 STORE C = &B LOAD X
938 STORE D = 4 LOAD C (gets &B)
939 LOAD *C (reads B)
940
941Without intervention, CPU 2 may perceive the events on CPU 1 in some
942effectively random order, despite the write barrier issued by CPU 1:
943
944 +-------+ : : : :
945 | | +------+ +-------+ | Sequence of update
946 | |------>| B=2 |----- --->| Y->8 | | of perception on
947 | | : +------+ \ +-------+ | CPU 2
948 | CPU 1 | : | A=1 | \ --->| C->&Y | V
949 | | +------+ | +-------+
950 | | wwwwwwwwwwwwwwww | : :
951 | | +------+ | : :
952 | | : | C=&B |--- | : : +-------+
953 | | : +------+ \ | +-------+ | |
954 | |------>| D=4 | ----------->| C->&B |------>| |
955 | | +------+ | +-------+ | |
956 +-------+ : : | : : | |
957 | : : | |
958 | : : | CPU 2 |
959 | +-------+ | |
960 Apparently incorrect ---> | | B->7 |------>| |
961 perception of B (!) | +-------+ | |
962 | : : | |
963 | +-------+ | |
964 The load of X holds ---> \ | X->9 |------>| |
965 up the maintenance \ +-------+ | |
966 of coherence of B ----->| B->2 | +-------+
967 +-------+
968 : :
969
970
971In the above example, CPU 2 perceives that B is 7, despite the load of *C
Paolo Ornati670e9f32006-10-03 22:57:56 +0200972(which would be B) coming after the LOAD of C.
David Howells108b42b2006-03-31 16:00:29 +0100973
974If, however, a data dependency barrier were to be placed between the load of C
David Howellsc14038c2006-04-10 22:54:24 -0700975and the load of *C (ie: B) on CPU 2:
976
977 CPU 1 CPU 2
978 ======================= =======================
979 { B = 7; X = 9; Y = 8; C = &Y }
980 STORE A = 1
981 STORE B = 2
982 <write barrier>
983 STORE C = &B LOAD X
984 STORE D = 4 LOAD C (gets &B)
985 <data dependency barrier>
986 LOAD *C (reads B)
987
988then the following will occur:
David Howells108b42b2006-03-31 16:00:29 +0100989
990 +-------+ : : : :
991 | | +------+ +-------+
992 | |------>| B=2 |----- --->| Y->8 |
993 | | : +------+ \ +-------+
994 | CPU 1 | : | A=1 | \ --->| C->&Y |
995 | | +------+ | +-------+
996 | | wwwwwwwwwwwwwwww | : :
997 | | +------+ | : :
998 | | : | C=&B |--- | : : +-------+
999 | | : +------+ \ | +-------+ | |
1000 | |------>| D=4 | ----------->| C->&B |------>| |
1001 | | +------+ | +-------+ | |
1002 +-------+ : : | : : | |
1003 | : : | |
1004 | : : | CPU 2 |
1005 | +-------+ | |
David Howells670bd952006-06-10 09:54:12 -07001006 | | X->9 |------>| |
1007 | +-------+ | |
1008 Makes sure all effects ---> \ ddddddddddddddddd | |
1009 prior to the store of C \ +-------+ | |
1010 are perceptible to ----->| B->2 |------>| |
1011 subsequent loads +-------+ | |
David Howells108b42b2006-03-31 16:00:29 +01001012 : : +-------+
1013
1014
1015And thirdly, a read barrier acts as a partial order on loads. Consider the
1016following sequence of events:
1017
1018 CPU 1 CPU 2
1019 ======================= =======================
David Howells670bd952006-06-10 09:54:12 -07001020 { A = 0, B = 9 }
David Howells108b42b2006-03-31 16:00:29 +01001021 STORE A=1
David Howells108b42b2006-03-31 16:00:29 +01001022 <write barrier>
David Howells670bd952006-06-10 09:54:12 -07001023 STORE B=2
David Howells108b42b2006-03-31 16:00:29 +01001024 LOAD B
David Howells670bd952006-06-10 09:54:12 -07001025 LOAD A
David Howells108b42b2006-03-31 16:00:29 +01001026
1027Without intervention, CPU 2 may then choose to perceive the events on CPU 1 in
1028some effectively random order, despite the write barrier issued by CPU 1:
1029
David Howells670bd952006-06-10 09:54:12 -07001030 +-------+ : : : :
1031 | | +------+ +-------+
1032 | |------>| A=1 |------ --->| A->0 |
1033 | | +------+ \ +-------+
1034 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1035 | | +------+ | +-------+
1036 | |------>| B=2 |--- | : :
1037 | | +------+ \ | : : +-------+
1038 +-------+ : : \ | +-------+ | |
1039 ---------->| B->2 |------>| |
1040 | +-------+ | CPU 2 |
1041 | | A->0 |------>| |
1042 | +-------+ | |
1043 | : : +-------+
1044 \ : :
1045 \ +-------+
1046 ---->| A->1 |
1047 +-------+
1048 : :
David Howells108b42b2006-03-31 16:00:29 +01001049
1050
David Howells6bc39272006-06-25 05:49:22 -07001051If, however, a read barrier were to be placed between the load of B and the
David Howells670bd952006-06-10 09:54:12 -07001052load of A on CPU 2:
David Howells108b42b2006-03-31 16:00:29 +01001053
David Howells670bd952006-06-10 09:54:12 -07001054 CPU 1 CPU 2
1055 ======================= =======================
1056 { A = 0, B = 9 }
1057 STORE A=1
1058 <write barrier>
1059 STORE B=2
1060 LOAD B
1061 <read barrier>
1062 LOAD A
1063
1064then the partial ordering imposed by CPU 1 will be perceived correctly by CPU
10652:
1066
1067 +-------+ : : : :
1068 | | +------+ +-------+
1069 | |------>| A=1 |------ --->| A->0 |
1070 | | +------+ \ +-------+
1071 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1072 | | +------+ | +-------+
1073 | |------>| B=2 |--- | : :
1074 | | +------+ \ | : : +-------+
1075 +-------+ : : \ | +-------+ | |
1076 ---------->| B->2 |------>| |
1077 | +-------+ | CPU 2 |
1078 | : : | |
1079 | : : | |
1080 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1081 barrier causes all effects \ +-------+ | |
1082 prior to the storage of B ---->| A->1 |------>| |
1083 to be perceptible to CPU 2 +-------+ | |
1084 : : +-------+
1085
1086
1087To illustrate this more completely, consider what could happen if the code
1088contained a load of A either side of the read barrier:
1089
1090 CPU 1 CPU 2
1091 ======================= =======================
1092 { A = 0, B = 9 }
1093 STORE A=1
1094 <write barrier>
1095 STORE B=2
1096 LOAD B
1097 LOAD A [first load of A]
1098 <read barrier>
1099 LOAD A [second load of A]
1100
1101Even though the two loads of A both occur after the load of B, they may both
1102come up with different values:
1103
1104 +-------+ : : : :
1105 | | +------+ +-------+
1106 | |------>| A=1 |------ --->| A->0 |
1107 | | +------+ \ +-------+
1108 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1109 | | +------+ | +-------+
1110 | |------>| B=2 |--- | : :
1111 | | +------+ \ | : : +-------+
1112 +-------+ : : \ | +-------+ | |
1113 ---------->| B->2 |------>| |
1114 | +-------+ | CPU 2 |
1115 | : : | |
1116 | : : | |
1117 | +-------+ | |
1118 | | A->0 |------>| 1st |
1119 | +-------+ | |
1120 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1121 barrier causes all effects \ +-------+ | |
1122 prior to the storage of B ---->| A->1 |------>| 2nd |
1123 to be perceptible to CPU 2 +-------+ | |
1124 : : +-------+
1125
1126
1127But it may be that the update to A from CPU 1 becomes perceptible to CPU 2
1128before the read barrier completes anyway:
1129
1130 +-------+ : : : :
1131 | | +------+ +-------+
1132 | |------>| A=1 |------ --->| A->0 |
1133 | | +------+ \ +-------+
1134 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1135 | | +------+ | +-------+
1136 | |------>| B=2 |--- | : :
1137 | | +------+ \ | : : +-------+
1138 +-------+ : : \ | +-------+ | |
1139 ---------->| B->2 |------>| |
1140 | +-------+ | CPU 2 |
1141 | : : | |
1142 \ : : | |
1143 \ +-------+ | |
1144 ---->| A->1 |------>| 1st |
1145 +-------+ | |
1146 rrrrrrrrrrrrrrrrr | |
1147 +-------+ | |
1148 | A->1 |------>| 2nd |
1149 +-------+ | |
1150 : : +-------+
1151
1152
1153The guarantee is that the second load will always come up with A == 1 if the
1154load of B came up with B == 2. No such guarantee exists for the first load of
1155A; that may come up with either A == 0 or A == 1.
1156
1157
1158READ MEMORY BARRIERS VS LOAD SPECULATION
1159----------------------------------------
1160
1161Many CPUs speculate with loads: that is they see that they will need to load an
1162item from memory, and they find a time where they're not using the bus for any
1163other loads, and so do the load in advance - even though they haven't actually
1164got to that point in the instruction execution flow yet. This permits the
1165actual load instruction to potentially complete immediately because the CPU
1166already has the value to hand.
1167
1168It may turn out that the CPU didn't actually need the value - perhaps because a
1169branch circumvented the load - in which case it can discard the value or just
1170cache it for later use.
1171
1172Consider:
1173
Ingo Molnare0edc782013-11-22 11:24:53 +01001174 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001175 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001176 LOAD B
1177 DIVIDE } Divide instructions generally
1178 DIVIDE } take a long time to perform
1179 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001180
1181Which might appear as this:
1182
1183 : : +-------+
1184 +-------+ | |
1185 --->| B->2 |------>| |
1186 +-------+ | CPU 2 |
1187 : :DIVIDE | |
1188 +-------+ | |
1189 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1190 division speculates on the +-------+ ~ | |
1191 LOAD of A : : ~ | |
1192 : :DIVIDE | |
1193 : : ~ | |
1194 Once the divisions are complete --> : : ~-->| |
1195 the CPU can then perform the : : | |
1196 LOAD with immediate effect : : +-------+
1197
1198
1199Placing a read barrier or a data dependency barrier just before the second
1200load:
1201
Ingo Molnare0edc782013-11-22 11:24:53 +01001202 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001203 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001204 LOAD B
1205 DIVIDE
1206 DIVIDE
David Howells670bd952006-06-10 09:54:12 -07001207 <read barrier>
Ingo Molnare0edc782013-11-22 11:24:53 +01001208 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001209
1210will force any value speculatively obtained to be reconsidered to an extent
1211dependent on the type of barrier used. If there was no change made to the
1212speculated memory location, then the speculated value will just be used:
1213
1214 : : +-------+
1215 +-------+ | |
1216 --->| B->2 |------>| |
1217 +-------+ | CPU 2 |
1218 : :DIVIDE | |
1219 +-------+ | |
1220 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1221 division speculates on the +-------+ ~ | |
1222 LOAD of A : : ~ | |
1223 : :DIVIDE | |
1224 : : ~ | |
1225 : : ~ | |
1226 rrrrrrrrrrrrrrrr~ | |
1227 : : ~ | |
1228 : : ~-->| |
1229 : : | |
1230 : : +-------+
1231
1232
1233but if there was an update or an invalidation from another CPU pending, then
1234the speculation will be cancelled and the value reloaded:
1235
1236 : : +-------+
1237 +-------+ | |
1238 --->| B->2 |------>| |
1239 +-------+ | CPU 2 |
1240 : :DIVIDE | |
1241 +-------+ | |
1242 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1243 division speculates on the +-------+ ~ | |
1244 LOAD of A : : ~ | |
1245 : :DIVIDE | |
1246 : : ~ | |
1247 : : ~ | |
1248 rrrrrrrrrrrrrrrrr | |
1249 +-------+ | |
1250 The speculation is discarded ---> --->| A->1 |------>| |
1251 and an updated value is +-------+ | |
1252 retrieved : : +-------+
David Howells108b42b2006-03-31 16:00:29 +01001253
1254
Paul E. McKenney241e6662011-02-10 16:54:50 -08001255TRANSITIVITY
1256------------
1257
1258Transitivity is a deeply intuitive notion about ordering that is not
1259always provided by real computer systems. The following example
1260demonstrates transitivity (also called "cumulativity"):
1261
1262 CPU 1 CPU 2 CPU 3
1263 ======================= ======================= =======================
1264 { X = 0, Y = 0 }
1265 STORE X=1 LOAD X STORE Y=1
1266 <general barrier> <general barrier>
1267 LOAD Y LOAD X
1268
1269Suppose that CPU 2's load from X returns 1 and its load from Y returns 0.
1270This indicates that CPU 2's load from X in some sense follows CPU 1's
1271store to X and that CPU 2's load from Y in some sense preceded CPU 3's
1272store to Y. The question is then "Can CPU 3's load from X return 0?"
1273
1274Because CPU 2's load from X in some sense came after CPU 1's store, it
1275is natural to expect that CPU 3's load from X must therefore return 1.
1276This expectation is an example of transitivity: if a load executing on
1277CPU A follows a load from the same variable executing on CPU B, then
1278CPU A's load must either return the same value that CPU B's load did,
1279or must return some later value.
1280
1281In the Linux kernel, use of general memory barriers guarantees
1282transitivity. Therefore, in the above example, if CPU 2's load from X
1283returns 1 and its load from Y returns 0, then CPU 3's load from X must
1284also return 1.
1285
1286However, transitivity is -not- guaranteed for read or write barriers.
1287For example, suppose that CPU 2's general barrier in the above example
1288is changed to a read barrier as shown below:
1289
1290 CPU 1 CPU 2 CPU 3
1291 ======================= ======================= =======================
1292 { X = 0, Y = 0 }
1293 STORE X=1 LOAD X STORE Y=1
1294 <read barrier> <general barrier>
1295 LOAD Y LOAD X
1296
1297This substitution destroys transitivity: in this example, it is perfectly
1298legal for CPU 2's load from X to return 1, its load from Y to return 0,
1299and CPU 3's load from X to return 0.
1300
1301The key point is that although CPU 2's read barrier orders its pair
1302of loads, it does not guarantee to order CPU 1's store. Therefore, if
1303this example runs on a system where CPUs 1 and 2 share a store buffer
1304or a level of cache, CPU 2 might have early access to CPU 1's writes.
1305General barriers are therefore required to ensure that all CPUs agree
1306on the combined order of CPU 1's and CPU 2's accesses.
1307
1308To reiterate, if your code requires transitivity, use general barriers
1309throughout.
1310
1311
David Howells108b42b2006-03-31 16:00:29 +01001312========================
1313EXPLICIT KERNEL BARRIERS
1314========================
1315
1316The Linux kernel has a variety of different barriers that act at different
1317levels:
1318
1319 (*) Compiler barrier.
1320
1321 (*) CPU memory barriers.
1322
1323 (*) MMIO write barrier.
1324
1325
1326COMPILER BARRIER
1327----------------
1328
1329The Linux kernel has an explicit compiler barrier function that prevents the
1330compiler from moving the memory accesses either side of it to the other side:
1331
1332 barrier();
1333
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001334This is a general barrier -- there are no read-read or write-write
1335variants of barrier(). However, READ_ONCE() and WRITE_ONCE() can be
1336thought of as weak forms of barrier() that affect only the specific
1337accesses flagged by the READ_ONCE() or WRITE_ONCE().
David Howells108b42b2006-03-31 16:00:29 +01001338
Paul E. McKenney692118d2013-12-11 13:59:07 -08001339The barrier() function has the following effects:
1340
1341 (*) Prevents the compiler from reordering accesses following the
1342 barrier() to precede any accesses preceding the barrier().
1343 One example use for this property is to ease communication between
1344 interrupt-handler code and the code that was interrupted.
1345
1346 (*) Within a loop, forces the compiler to load the variables used
1347 in that loop's conditional on each pass through that loop.
1348
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001349The READ_ONCE() and WRITE_ONCE() functions can prevent any number of
1350optimizations that, while perfectly safe in single-threaded code, can
1351be fatal in concurrent code. Here are some examples of these sorts
1352of optimizations:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001353
Paul E. McKenney449f7412014-01-02 15:03:50 -08001354 (*) The compiler is within its rights to reorder loads and stores
1355 to the same variable, and in some cases, the CPU is within its
1356 rights to reorder loads to the same variable. This means that
1357 the following code:
1358
1359 a[0] = x;
1360 a[1] = x;
1361
1362 Might result in an older value of x stored in a[1] than in a[0].
1363 Prevent both the compiler and the CPU from doing this as follows:
1364
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001365 a[0] = READ_ONCE(x);
1366 a[1] = READ_ONCE(x);
Paul E. McKenney449f7412014-01-02 15:03:50 -08001367
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001368 In short, READ_ONCE() and WRITE_ONCE() provide cache coherence for
1369 accesses from multiple CPUs to a single variable.
Paul E. McKenney449f7412014-01-02 15:03:50 -08001370
Paul E. McKenney692118d2013-12-11 13:59:07 -08001371 (*) The compiler is within its rights to merge successive loads from
1372 the same variable. Such merging can cause the compiler to "optimize"
1373 the following code:
1374
1375 while (tmp = a)
1376 do_something_with(tmp);
1377
1378 into the following code, which, although in some sense legitimate
1379 for single-threaded code, is almost certainly not what the developer
1380 intended:
1381
1382 if (tmp = a)
1383 for (;;)
1384 do_something_with(tmp);
1385
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001386 Use READ_ONCE() to prevent the compiler from doing this to you:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001387
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001388 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001389 do_something_with(tmp);
1390
1391 (*) The compiler is within its rights to reload a variable, for example,
1392 in cases where high register pressure prevents the compiler from
1393 keeping all data of interest in registers. The compiler might
1394 therefore optimize the variable 'tmp' out of our previous example:
1395
1396 while (tmp = a)
1397 do_something_with(tmp);
1398
1399 This could result in the following code, which is perfectly safe in
1400 single-threaded code, but can be fatal in concurrent code:
1401
1402 while (a)
1403 do_something_with(a);
1404
1405 For example, the optimized version of this code could result in
1406 passing a zero to do_something_with() in the case where the variable
1407 a was modified by some other CPU between the "while" statement and
1408 the call to do_something_with().
1409
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001410 Again, use READ_ONCE() to prevent the compiler from doing this:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001411
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001412 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001413 do_something_with(tmp);
1414
1415 Note that if the compiler runs short of registers, it might save
1416 tmp onto the stack. The overhead of this saving and later restoring
1417 is why compilers reload variables. Doing so is perfectly safe for
1418 single-threaded code, so you need to tell the compiler about cases
1419 where it is not safe.
1420
1421 (*) The compiler is within its rights to omit a load entirely if it knows
1422 what the value will be. For example, if the compiler can prove that
1423 the value of variable 'a' is always zero, it can optimize this code:
1424
1425 while (tmp = a)
1426 do_something_with(tmp);
1427
1428 Into this:
1429
1430 do { } while (0);
1431
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001432 This transformation is a win for single-threaded code because it
1433 gets rid of a load and a branch. The problem is that the compiler
1434 will carry out its proof assuming that the current CPU is the only
1435 one updating variable 'a'. If variable 'a' is shared, then the
1436 compiler's proof will be erroneous. Use READ_ONCE() to tell the
1437 compiler that it doesn't know as much as it thinks it does:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001438
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001439 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001440 do_something_with(tmp);
1441
1442 But please note that the compiler is also closely watching what you
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001443 do with the value after the READ_ONCE(). For example, suppose you
Paul E. McKenney692118d2013-12-11 13:59:07 -08001444 do the following and MAX is a preprocessor macro with the value 1:
1445
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001446 while ((tmp = READ_ONCE(a)) % MAX)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001447 do_something_with(tmp);
1448
1449 Then the compiler knows that the result of the "%" operator applied
1450 to MAX will always be zero, again allowing the compiler to optimize
1451 the code into near-nonexistence. (It will still load from the
1452 variable 'a'.)
1453
1454 (*) Similarly, the compiler is within its rights to omit a store entirely
1455 if it knows that the variable already has the value being stored.
1456 Again, the compiler assumes that the current CPU is the only one
1457 storing into the variable, which can cause the compiler to do the
1458 wrong thing for shared variables. For example, suppose you have
1459 the following:
1460
1461 a = 0;
1462 /* Code that does not store to variable a. */
1463 a = 0;
1464
1465 The compiler sees that the value of variable 'a' is already zero, so
1466 it might well omit the second store. This would come as a fatal
1467 surprise if some other CPU might have stored to variable 'a' in the
1468 meantime.
1469
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001470 Use WRITE_ONCE() to prevent the compiler from making this sort of
Paul E. McKenney692118d2013-12-11 13:59:07 -08001471 wrong guess:
1472
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001473 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001474 /* Code that does not store to variable a. */
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001475 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001476
1477 (*) The compiler is within its rights to reorder memory accesses unless
1478 you tell it not to. For example, consider the following interaction
1479 between process-level code and an interrupt handler:
1480
1481 void process_level(void)
1482 {
1483 msg = get_message();
1484 flag = true;
1485 }
1486
1487 void interrupt_handler(void)
1488 {
1489 if (flag)
1490 process_message(msg);
1491 }
1492
Masanari Iidadf5cbb22014-03-21 10:04:30 +09001493 There is nothing to prevent the compiler from transforming
Paul E. McKenney692118d2013-12-11 13:59:07 -08001494 process_level() to the following, in fact, this might well be a
1495 win for single-threaded code:
1496
1497 void process_level(void)
1498 {
1499 flag = true;
1500 msg = get_message();
1501 }
1502
1503 If the interrupt occurs between these two statement, then
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001504 interrupt_handler() might be passed a garbled msg. Use WRITE_ONCE()
Paul E. McKenney692118d2013-12-11 13:59:07 -08001505 to prevent this as follows:
1506
1507 void process_level(void)
1508 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001509 WRITE_ONCE(msg, get_message());
1510 WRITE_ONCE(flag, true);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001511 }
1512
1513 void interrupt_handler(void)
1514 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001515 if (READ_ONCE(flag))
1516 process_message(READ_ONCE(msg));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001517 }
1518
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001519 Note that the READ_ONCE() and WRITE_ONCE() wrappers in
1520 interrupt_handler() are needed if this interrupt handler can itself
1521 be interrupted by something that also accesses 'flag' and 'msg',
1522 for example, a nested interrupt or an NMI. Otherwise, READ_ONCE()
1523 and WRITE_ONCE() are not needed in interrupt_handler() other than
1524 for documentation purposes. (Note also that nested interrupts
1525 do not typically occur in modern Linux kernels, in fact, if an
1526 interrupt handler returns with interrupts enabled, you will get a
1527 WARN_ONCE() splat.)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001528
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001529 You should assume that the compiler can move READ_ONCE() and
1530 WRITE_ONCE() past code not containing READ_ONCE(), WRITE_ONCE(),
1531 barrier(), or similar primitives.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001532
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001533 This effect could also be achieved using barrier(), but READ_ONCE()
1534 and WRITE_ONCE() are more selective: With READ_ONCE() and
1535 WRITE_ONCE(), the compiler need only forget the contents of the
1536 indicated memory locations, while with barrier() the compiler must
1537 discard the value of all memory locations that it has currented
1538 cached in any machine registers. Of course, the compiler must also
1539 respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
1540 though the CPU of course need not do so.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001541
1542 (*) The compiler is within its rights to invent stores to a variable,
1543 as in the following example:
1544
1545 if (a)
1546 b = a;
1547 else
1548 b = 42;
1549
1550 The compiler might save a branch by optimizing this as follows:
1551
1552 b = 42;
1553 if (a)
1554 b = a;
1555
1556 In single-threaded code, this is not only safe, but also saves
1557 a branch. Unfortunately, in concurrent code, this optimization
1558 could cause some other CPU to see a spurious value of 42 -- even
1559 if variable 'a' was never zero -- when loading variable 'b'.
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001560 Use WRITE_ONCE() to prevent this as follows:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001561
1562 if (a)
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001563 WRITE_ONCE(b, a);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001564 else
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001565 WRITE_ONCE(b, 42);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001566
1567 The compiler can also invent loads. These are usually less
1568 damaging, but they can result in cache-line bouncing and thus in
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001569 poor performance and scalability. Use READ_ONCE() to prevent
Paul E. McKenney692118d2013-12-11 13:59:07 -08001570 invented loads.
1571
1572 (*) For aligned memory locations whose size allows them to be accessed
1573 with a single memory-reference instruction, prevents "load tearing"
1574 and "store tearing," in which a single large access is replaced by
1575 multiple smaller accesses. For example, given an architecture having
1576 16-bit store instructions with 7-bit immediate fields, the compiler
1577 might be tempted to use two 16-bit store-immediate instructions to
1578 implement the following 32-bit store:
1579
1580 p = 0x00010002;
1581
1582 Please note that GCC really does use this sort of optimization,
1583 which is not surprising given that it would likely take more
1584 than two instructions to build the constant and then store it.
1585 This optimization can therefore be a win in single-threaded code.
1586 In fact, a recent bug (since fixed) caused GCC to incorrectly use
1587 this optimization in a volatile store. In the absence of such bugs,
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001588 use of WRITE_ONCE() prevents store tearing in the following example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001589
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001590 WRITE_ONCE(p, 0x00010002);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001591
1592 Use of packed structures can also result in load and store tearing,
1593 as in this example:
1594
1595 struct __attribute__((__packed__)) foo {
1596 short a;
1597 int b;
1598 short c;
1599 };
1600 struct foo foo1, foo2;
1601 ...
1602
1603 foo2.a = foo1.a;
1604 foo2.b = foo1.b;
1605 foo2.c = foo1.c;
1606
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001607 Because there are no READ_ONCE() or WRITE_ONCE() wrappers and no
1608 volatile markings, the compiler would be well within its rights to
1609 implement these three assignment statements as a pair of 32-bit
1610 loads followed by a pair of 32-bit stores. This would result in
1611 load tearing on 'foo1.b' and store tearing on 'foo2.b'. READ_ONCE()
1612 and WRITE_ONCE() again prevent tearing in this example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001613
1614 foo2.a = foo1.a;
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001615 WRITE_ONCE(foo2.b, READ_ONCE(foo1.b));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001616 foo2.c = foo1.c;
1617
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001618All that aside, it is never necessary to use READ_ONCE() and
1619WRITE_ONCE() on a variable that has been marked volatile. For example,
1620because 'jiffies' is marked volatile, it is never necessary to
1621say READ_ONCE(jiffies). The reason for this is that READ_ONCE() and
1622WRITE_ONCE() are implemented as volatile casts, which has no effect when
1623its argument is already marked volatile.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001624
1625Please note that these compiler barriers have no direct effect on the CPU,
1626which may then reorder things however it wishes.
David Howells108b42b2006-03-31 16:00:29 +01001627
1628
1629CPU MEMORY BARRIERS
1630-------------------
1631
1632The Linux kernel has eight basic CPU memory barriers:
1633
1634 TYPE MANDATORY SMP CONDITIONAL
1635 =============== ======================= ===========================
1636 GENERAL mb() smp_mb()
1637 WRITE wmb() smp_wmb()
1638 READ rmb() smp_rmb()
1639 DATA DEPENDENCY read_barrier_depends() smp_read_barrier_depends()
1640
1641
Nick Piggin73f10282008-05-14 06:35:11 +02001642All memory barriers except the data dependency barriers imply a compiler
1643barrier. Data dependencies do not impose any additional compiler ordering.
1644
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001645Aside: In the case of data dependencies, the compiler would be expected
1646to issue the loads in the correct order (eg. `a[b]` would have to load
1647the value of b before loading a[b]), however there is no guarantee in
1648the C specification that the compiler may not speculate the value of b
1649(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
1650tmp = a[b]; ). There is also the problem of a compiler reloading b after
1651having loaded a[b], thus having a newer copy of b than a[b]. A consensus
1652has not yet been reached about these problems, however the READ_ONCE()
1653macro is a good place to start looking.
David Howells108b42b2006-03-31 16:00:29 +01001654
1655SMP memory barriers are reduced to compiler barriers on uniprocessor compiled
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001656systems because it is assumed that a CPU will appear to be self-consistent,
David Howells108b42b2006-03-31 16:00:29 +01001657and will order overlapping accesses correctly with respect to itself.
1658
1659[!] Note that SMP memory barriers _must_ be used to control the ordering of
1660references to shared memory on SMP systems, though the use of locking instead
1661is sufficient.
1662
1663Mandatory barriers should not be used to control SMP effects, since mandatory
1664barriers unnecessarily impose overhead on UP systems. They may, however, be
1665used to control MMIO effects on accesses through relaxed memory I/O windows.
1666These are required even on non-SMP systems as they affect the order in which
1667memory operations appear to a device by prohibiting both the compiler and the
1668CPU from reordering them.
1669
1670
1671There are some more advanced barrier functions:
1672
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001673 (*) smp_store_mb(var, value)
David Howells108b42b2006-03-31 16:00:29 +01001674
Oleg Nesterov75b2bd52006-11-08 17:44:38 -08001675 This assigns the value to the variable and then inserts a full memory
Steven Rostedtf92213b2006-07-14 16:05:01 -04001676 barrier after it, depending on the function. It isn't guaranteed to
David Howells108b42b2006-03-31 16:00:29 +01001677 insert anything more than a compiler barrier in a UP compilation.
1678
1679
Peter Zijlstra1b156112014-03-13 19:00:35 +01001680 (*) smp_mb__before_atomic();
1681 (*) smp_mb__after_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001682
Peter Zijlstra1b156112014-03-13 19:00:35 +01001683 These are for use with atomic (such as add, subtract, increment and
1684 decrement) functions that don't return a value, especially when used for
1685 reference counting. These functions do not imply memory barriers.
1686
1687 These are also used for atomic bitop functions that do not return a
1688 value (such as set_bit and clear_bit).
David Howells108b42b2006-03-31 16:00:29 +01001689
1690 As an example, consider a piece of code that marks an object as being dead
1691 and then decrements the object's reference count:
1692
1693 obj->dead = 1;
Peter Zijlstra1b156112014-03-13 19:00:35 +01001694 smp_mb__before_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001695 atomic_dec(&obj->ref_count);
1696
1697 This makes sure that the death mark on the object is perceived to be set
1698 *before* the reference counter is decremented.
1699
1700 See Documentation/atomic_ops.txt for more information. See the "Atomic
1701 operations" subsection for information on where to use these.
1702
1703
Paul E. McKenneyad2ad5d2015-09-17 08:18:32 -07001704 (*) lockless_dereference();
1705 This can be thought of as a pointer-fetch wrapper around the
1706 smp_read_barrier_depends() data-dependency barrier.
1707
1708 This is also similar to rcu_dereference(), but in cases where
1709 object lifetime is handled by some mechanism other than RCU, for
1710 example, when the objects removed only when the system goes down.
1711 In addition, lockless_dereference() is used in some data structures
1712 that can be used both with and without RCU.
1713
1714
Alexander Duyck1077fa32014-12-11 15:02:06 -08001715 (*) dma_wmb();
1716 (*) dma_rmb();
1717
1718 These are for use with consistent memory to guarantee the ordering
1719 of writes or reads of shared memory accessible to both the CPU and a
1720 DMA capable device.
1721
1722 For example, consider a device driver that shares memory with a device
1723 and uses a descriptor status value to indicate if the descriptor belongs
1724 to the device or the CPU, and a doorbell to notify it when new
1725 descriptors are available:
1726
1727 if (desc->status != DEVICE_OWN) {
1728 /* do not read data until we own descriptor */
1729 dma_rmb();
1730
1731 /* read/modify data */
1732 read_data = desc->data;
1733 desc->data = write_data;
1734
1735 /* flush modifications before status update */
1736 dma_wmb();
1737
1738 /* assign ownership */
1739 desc->status = DEVICE_OWN;
1740
1741 /* force memory to sync before notifying device via MMIO */
1742 wmb();
1743
1744 /* notify device of new descriptors */
1745 writel(DESC_NOTIFY, doorbell);
1746 }
1747
1748 The dma_rmb() allows us guarantee the device has released ownership
Sylvain Trias7a458002015-04-08 10:27:57 +02001749 before we read the data from the descriptor, and the dma_wmb() allows
Alexander Duyck1077fa32014-12-11 15:02:06 -08001750 us to guarantee the data is written to the descriptor before the device
1751 can see it now has ownership. The wmb() is needed to guarantee that the
1752 cache coherent memory writes have completed before attempting a write to
1753 the cache incoherent MMIO region.
1754
1755 See Documentation/DMA-API.txt for more information on consistent memory.
1756
David Howells108b42b2006-03-31 16:00:29 +01001757MMIO WRITE BARRIER
1758------------------
1759
1760The Linux kernel also has a special barrier for use with memory-mapped I/O
1761writes:
1762
1763 mmiowb();
1764
1765This is a variation on the mandatory write barrier that causes writes to weakly
1766ordered I/O regions to be partially ordered. Its effects may go beyond the
1767CPU->Hardware interface and actually affect the hardware at some level.
1768
1769See the subsection "Locks vs I/O accesses" for more information.
1770
1771
1772===============================
1773IMPLICIT KERNEL MEMORY BARRIERS
1774===============================
1775
1776Some of the other functions in the linux kernel imply memory barriers, amongst
David Howells670bd952006-06-10 09:54:12 -07001777which are locking and scheduling functions.
David Howells108b42b2006-03-31 16:00:29 +01001778
1779This specification is a _minimum_ guarantee; any particular architecture may
1780provide more substantial guarantees, but these may not be relied upon outside
1781of arch specific code.
1782
1783
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001784ACQUIRING FUNCTIONS
1785-------------------
David Howells108b42b2006-03-31 16:00:29 +01001786
1787The Linux kernel has a number of locking constructs:
1788
1789 (*) spin locks
1790 (*) R/W spin locks
1791 (*) mutexes
1792 (*) semaphores
1793 (*) R/W semaphores
David Howells108b42b2006-03-31 16:00:29 +01001794
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001795In all cases there are variants on "ACQUIRE" operations and "RELEASE" operations
David Howells108b42b2006-03-31 16:00:29 +01001796for each construct. These operations all imply certain barriers:
1797
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001798 (1) ACQUIRE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001799
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001800 Memory operations issued after the ACQUIRE will be completed after the
1801 ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001802
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001803 Memory operations issued before the ACQUIRE may be completed after
1804 the ACQUIRE operation has completed. An smp_mb__before_spinlock(),
Will Deacond9560282015-03-31 09:39:41 +01001805 combined with a following ACQUIRE, orders prior stores against
1806 subsequent loads and stores. Note that this is weaker than smp_mb()!
1807 The smp_mb__before_spinlock() primitive is free on many architectures.
David Howells108b42b2006-03-31 16:00:29 +01001808
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001809 (2) RELEASE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001810
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001811 Memory operations issued before the RELEASE will be completed before the
1812 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001813
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001814 Memory operations issued after the RELEASE may be completed before the
1815 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001816
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001817 (3) ACQUIRE vs ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001818
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001819 All ACQUIRE operations issued before another ACQUIRE operation will be
1820 completed before that ACQUIRE operation.
David Howells108b42b2006-03-31 16:00:29 +01001821
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001822 (4) ACQUIRE vs RELEASE implication:
David Howells108b42b2006-03-31 16:00:29 +01001823
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001824 All ACQUIRE operations issued before a RELEASE operation will be
1825 completed before the RELEASE operation.
David Howells108b42b2006-03-31 16:00:29 +01001826
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001827 (5) Failed conditional ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001828
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001829 Certain locking variants of the ACQUIRE operation may fail, either due to
1830 being unable to get the lock immediately, or due to receiving an unblocked
David Howells108b42b2006-03-31 16:00:29 +01001831 signal whilst asleep waiting for the lock to become available. Failed
1832 locks do not imply any sort of barrier.
1833
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001834[!] Note: one of the consequences of lock ACQUIREs and RELEASEs being only
1835one-way barriers is that the effects of instructions outside of a critical
1836section may seep into the inside of the critical section.
David Howells108b42b2006-03-31 16:00:29 +01001837
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001838An ACQUIRE followed by a RELEASE may not be assumed to be full memory barrier
1839because it is possible for an access preceding the ACQUIRE to happen after the
1840ACQUIRE, and an access following the RELEASE to happen before the RELEASE, and
1841the two accesses can themselves then cross:
David Howells670bd952006-06-10 09:54:12 -07001842
1843 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001844 ACQUIRE M
1845 RELEASE M
David Howells670bd952006-06-10 09:54:12 -07001846 *B = b;
1847
1848may occur as:
1849
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001850 ACQUIRE M, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001851
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001852When the ACQUIRE and RELEASE are a lock acquisition and release,
1853respectively, this same reordering can occur if the lock's ACQUIRE and
1854RELEASE are to the same lock variable, but only from the perspective of
1855another CPU not holding that lock. In short, a ACQUIRE followed by an
1856RELEASE may -not- be assumed to be a full memory barrier.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001857
Paul E. McKenney12d560f2015-07-14 18:35:23 -07001858Similarly, the reverse case of a RELEASE followed by an ACQUIRE does
1859not imply a full memory barrier. Therefore, the CPU's execution of the
1860critical sections corresponding to the RELEASE and the ACQUIRE can cross,
1861so that:
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001862
1863 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001864 RELEASE M
1865 ACQUIRE N
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001866 *B = b;
1867
1868could occur as:
1869
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001870 ACQUIRE N, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001871
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001872It might appear that this reordering could introduce a deadlock.
1873However, this cannot happen because if such a deadlock threatened,
1874the RELEASE would simply complete, thereby avoiding the deadlock.
1875
1876 Why does this work?
1877
1878 One key point is that we are only talking about the CPU doing
1879 the reordering, not the compiler. If the compiler (or, for
1880 that matter, the developer) switched the operations, deadlock
1881 -could- occur.
1882
1883 But suppose the CPU reordered the operations. In this case,
1884 the unlock precedes the lock in the assembly code. The CPU
1885 simply elected to try executing the later lock operation first.
1886 If there is a deadlock, this lock operation will simply spin (or
1887 try to sleep, but more on that later). The CPU will eventually
1888 execute the unlock operation (which preceded the lock operation
1889 in the assembly code), which will unravel the potential deadlock,
1890 allowing the lock operation to succeed.
1891
1892 But what if the lock is a sleeplock? In that case, the code will
1893 try to enter the scheduler, where it will eventually encounter
1894 a memory barrier, which will force the earlier unlock operation
1895 to complete, again unraveling the deadlock. There might be
1896 a sleep-unlock race, but the locking primitive needs to resolve
1897 such races properly in any case.
1898
David Howells108b42b2006-03-31 16:00:29 +01001899Locks and semaphores may not provide any guarantee of ordering on UP compiled
1900systems, and so cannot be counted on in such a situation to actually achieve
1901anything at all - especially with respect to I/O accesses - unless combined
1902with interrupt disabling operations.
1903
1904See also the section on "Inter-CPU locking barrier effects".
1905
1906
1907As an example, consider the following:
1908
1909 *A = a;
1910 *B = b;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001911 ACQUIRE
David Howells108b42b2006-03-31 16:00:29 +01001912 *C = c;
1913 *D = d;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001914 RELEASE
David Howells108b42b2006-03-31 16:00:29 +01001915 *E = e;
1916 *F = f;
1917
1918The following sequence of events is acceptable:
1919
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001920 ACQUIRE, {*F,*A}, *E, {*C,*D}, *B, RELEASE
David Howells108b42b2006-03-31 16:00:29 +01001921
1922 [+] Note that {*F,*A} indicates a combined access.
1923
1924But none of the following are:
1925
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001926 {*F,*A}, *B, ACQUIRE, *C, *D, RELEASE, *E
1927 *A, *B, *C, ACQUIRE, *D, RELEASE, *E, *F
1928 *A, *B, ACQUIRE, *C, RELEASE, *D, *E, *F
1929 *B, ACQUIRE, *C, *D, RELEASE, {*F,*A}, *E
David Howells108b42b2006-03-31 16:00:29 +01001930
1931
1932
1933INTERRUPT DISABLING FUNCTIONS
1934-----------------------------
1935
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001936Functions that disable interrupts (ACQUIRE equivalent) and enable interrupts
1937(RELEASE equivalent) will act as compiler barriers only. So if memory or I/O
David Howells108b42b2006-03-31 16:00:29 +01001938barriers are required in such a situation, they must be provided from some
1939other means.
1940
1941
David Howells50fa6102009-04-28 15:01:38 +01001942SLEEP AND WAKE-UP FUNCTIONS
1943---------------------------
1944
1945Sleeping and waking on an event flagged in global data can be viewed as an
1946interaction between two pieces of data: the task state of the task waiting for
1947the event and the global data used to indicate the event. To make sure that
1948these appear to happen in the right order, the primitives to begin the process
1949of going to sleep, and the primitives to initiate a wake up imply certain
1950barriers.
1951
1952Firstly, the sleeper normally follows something like this sequence of events:
1953
1954 for (;;) {
1955 set_current_state(TASK_UNINTERRUPTIBLE);
1956 if (event_indicated)
1957 break;
1958 schedule();
1959 }
1960
1961A general memory barrier is interpolated automatically by set_current_state()
1962after it has altered the task state:
1963
1964 CPU 1
1965 ===============================
1966 set_current_state();
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001967 smp_store_mb();
David Howells50fa6102009-04-28 15:01:38 +01001968 STORE current->state
1969 <general barrier>
1970 LOAD event_indicated
1971
1972set_current_state() may be wrapped by:
1973
1974 prepare_to_wait();
1975 prepare_to_wait_exclusive();
1976
1977which therefore also imply a general memory barrier after setting the state.
1978The whole sequence above is available in various canned forms, all of which
1979interpolate the memory barrier in the right place:
1980
1981 wait_event();
1982 wait_event_interruptible();
1983 wait_event_interruptible_exclusive();
1984 wait_event_interruptible_timeout();
1985 wait_event_killable();
1986 wait_event_timeout();
1987 wait_on_bit();
1988 wait_on_bit_lock();
1989
1990
1991Secondly, code that performs a wake up normally follows something like this:
1992
1993 event_indicated = 1;
1994 wake_up(&event_wait_queue);
1995
1996or:
1997
1998 event_indicated = 1;
1999 wake_up_process(event_daemon);
2000
2001A write memory barrier is implied by wake_up() and co. if and only if they wake
2002something up. The barrier occurs before the task state is cleared, and so sits
2003between the STORE to indicate the event and the STORE to set TASK_RUNNING:
2004
2005 CPU 1 CPU 2
2006 =============================== ===============================
2007 set_current_state(); STORE event_indicated
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002008 smp_store_mb(); wake_up();
David Howells50fa6102009-04-28 15:01:38 +01002009 STORE current->state <write barrier>
2010 <general barrier> STORE current->state
2011 LOAD event_indicated
2012
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002013To repeat, this write memory barrier is present if and only if something
2014is actually awakened. To see this, consider the following sequence of
2015events, where X and Y are both initially zero:
2016
2017 CPU 1 CPU 2
2018 =============================== ===============================
2019 X = 1; STORE event_indicated
2020 smp_mb(); wake_up();
2021 Y = 1; wait_event(wq, Y == 1);
2022 wake_up(); load from Y sees 1, no memory barrier
2023 load from X might see 0
2024
2025In contrast, if a wakeup does occur, CPU 2's load from X would be guaranteed
2026to see 1.
2027
David Howells50fa6102009-04-28 15:01:38 +01002028The available waker functions include:
2029
2030 complete();
2031 wake_up();
2032 wake_up_all();
2033 wake_up_bit();
2034 wake_up_interruptible();
2035 wake_up_interruptible_all();
2036 wake_up_interruptible_nr();
2037 wake_up_interruptible_poll();
2038 wake_up_interruptible_sync();
2039 wake_up_interruptible_sync_poll();
2040 wake_up_locked();
2041 wake_up_locked_poll();
2042 wake_up_nr();
2043 wake_up_poll();
2044 wake_up_process();
2045
2046
2047[!] Note that the memory barriers implied by the sleeper and the waker do _not_
2048order multiple stores before the wake-up with respect to loads of those stored
2049values after the sleeper has called set_current_state(). For instance, if the
2050sleeper does:
2051
2052 set_current_state(TASK_INTERRUPTIBLE);
2053 if (event_indicated)
2054 break;
2055 __set_current_state(TASK_RUNNING);
2056 do_something(my_data);
2057
2058and the waker does:
2059
2060 my_data = value;
2061 event_indicated = 1;
2062 wake_up(&event_wait_queue);
2063
2064there's no guarantee that the change to event_indicated will be perceived by
2065the sleeper as coming after the change to my_data. In such a circumstance, the
2066code on both sides must interpolate its own memory barriers between the
2067separate data accesses. Thus the above sleeper ought to do:
2068
2069 set_current_state(TASK_INTERRUPTIBLE);
2070 if (event_indicated) {
2071 smp_rmb();
2072 do_something(my_data);
2073 }
2074
2075and the waker should do:
2076
2077 my_data = value;
2078 smp_wmb();
2079 event_indicated = 1;
2080 wake_up(&event_wait_queue);
2081
2082
David Howells108b42b2006-03-31 16:00:29 +01002083MISCELLANEOUS FUNCTIONS
2084-----------------------
2085
2086Other functions that imply barriers:
2087
2088 (*) schedule() and similar imply full memory barriers.
2089
David Howells108b42b2006-03-31 16:00:29 +01002090
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002091===================================
2092INTER-CPU ACQUIRING BARRIER EFFECTS
2093===================================
David Howells108b42b2006-03-31 16:00:29 +01002094
2095On SMP systems locking primitives give a more substantial form of barrier: one
2096that does affect memory access ordering on other CPUs, within the context of
2097conflict on any particular lock.
2098
2099
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002100ACQUIRES VS MEMORY ACCESSES
2101---------------------------
David Howells108b42b2006-03-31 16:00:29 +01002102
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002103Consider the following: the system has a pair of spinlocks (M) and (Q), and
David Howells108b42b2006-03-31 16:00:29 +01002104three CPUs; then should the following sequence of events occur:
2105
2106 CPU 1 CPU 2
2107 =============================== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002108 WRITE_ONCE(*A, a); WRITE_ONCE(*E, e);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002109 ACQUIRE M ACQUIRE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002110 WRITE_ONCE(*B, b); WRITE_ONCE(*F, f);
2111 WRITE_ONCE(*C, c); WRITE_ONCE(*G, g);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002112 RELEASE M RELEASE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002113 WRITE_ONCE(*D, d); WRITE_ONCE(*H, h);
David Howells108b42b2006-03-31 16:00:29 +01002114
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002115Then there is no guarantee as to what order CPU 3 will see the accesses to *A
David Howells108b42b2006-03-31 16:00:29 +01002116through *H occur in, other than the constraints imposed by the separate locks
2117on the separate CPUs. It might, for example, see:
2118
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002119 *E, ACQUIRE M, ACQUIRE Q, *G, *C, *F, *A, *B, RELEASE Q, *D, *H, RELEASE M
David Howells108b42b2006-03-31 16:00:29 +01002120
2121But it won't see any of:
2122
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002123 *B, *C or *D preceding ACQUIRE M
2124 *A, *B or *C following RELEASE M
2125 *F, *G or *H preceding ACQUIRE Q
2126 *E, *F or *G following RELEASE Q
David Howells108b42b2006-03-31 16:00:29 +01002127
2128
David Howells108b42b2006-03-31 16:00:29 +01002129
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002130ACQUIRES VS I/O ACCESSES
2131------------------------
David Howells108b42b2006-03-31 16:00:29 +01002132
2133Under certain circumstances (especially involving NUMA), I/O accesses within
2134two spinlocked sections on two different CPUs may be seen as interleaved by the
2135PCI bridge, because the PCI bridge does not necessarily participate in the
2136cache-coherence protocol, and is therefore incapable of issuing the required
2137read memory barriers.
2138
2139For example:
2140
2141 CPU 1 CPU 2
2142 =============================== ===============================
2143 spin_lock(Q)
2144 writel(0, ADDR)
2145 writel(1, DATA);
2146 spin_unlock(Q);
2147 spin_lock(Q);
2148 writel(4, ADDR);
2149 writel(5, DATA);
2150 spin_unlock(Q);
2151
2152may be seen by the PCI bridge as follows:
2153
2154 STORE *ADDR = 0, STORE *ADDR = 4, STORE *DATA = 1, STORE *DATA = 5
2155
2156which would probably cause the hardware to malfunction.
2157
2158
2159What is necessary here is to intervene with an mmiowb() before dropping the
2160spinlock, for example:
2161
2162 CPU 1 CPU 2
2163 =============================== ===============================
2164 spin_lock(Q)
2165 writel(0, ADDR)
2166 writel(1, DATA);
2167 mmiowb();
2168 spin_unlock(Q);
2169 spin_lock(Q);
2170 writel(4, ADDR);
2171 writel(5, DATA);
2172 mmiowb();
2173 spin_unlock(Q);
2174
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002175this will ensure that the two stores issued on CPU 1 appear at the PCI bridge
2176before either of the stores issued on CPU 2.
David Howells108b42b2006-03-31 16:00:29 +01002177
2178
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002179Furthermore, following a store by a load from the same device obviates the need
2180for the mmiowb(), because the load forces the store to complete before the load
David Howells108b42b2006-03-31 16:00:29 +01002181is performed:
2182
2183 CPU 1 CPU 2
2184 =============================== ===============================
2185 spin_lock(Q)
2186 writel(0, ADDR)
2187 a = readl(DATA);
2188 spin_unlock(Q);
2189 spin_lock(Q);
2190 writel(4, ADDR);
2191 b = readl(DATA);
2192 spin_unlock(Q);
2193
2194
2195See Documentation/DocBook/deviceiobook.tmpl for more information.
2196
2197
2198=================================
2199WHERE ARE MEMORY BARRIERS NEEDED?
2200=================================
2201
2202Under normal operation, memory operation reordering is generally not going to
2203be a problem as a single-threaded linear piece of code will still appear to
David Howells50fa6102009-04-28 15:01:38 +01002204work correctly, even if it's in an SMP kernel. There are, however, four
David Howells108b42b2006-03-31 16:00:29 +01002205circumstances in which reordering definitely _could_ be a problem:
2206
2207 (*) Interprocessor interaction.
2208
2209 (*) Atomic operations.
2210
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002211 (*) Accessing devices.
David Howells108b42b2006-03-31 16:00:29 +01002212
2213 (*) Interrupts.
2214
2215
2216INTERPROCESSOR INTERACTION
2217--------------------------
2218
2219When there's a system with more than one processor, more than one CPU in the
2220system may be working on the same data set at the same time. This can cause
2221synchronisation problems, and the usual way of dealing with them is to use
2222locks. Locks, however, are quite expensive, and so it may be preferable to
2223operate without the use of a lock if at all possible. In such a case
2224operations that affect both CPUs may have to be carefully ordered to prevent
2225a malfunction.
2226
2227Consider, for example, the R/W semaphore slow path. Here a waiting process is
2228queued on the semaphore, by virtue of it having a piece of its stack linked to
2229the semaphore's list of waiting processes:
2230
2231 struct rw_semaphore {
2232 ...
2233 spinlock_t lock;
2234 struct list_head waiters;
2235 };
2236
2237 struct rwsem_waiter {
2238 struct list_head list;
2239 struct task_struct *task;
2240 };
2241
2242To wake up a particular waiter, the up_read() or up_write() functions have to:
2243
2244 (1) read the next pointer from this waiter's record to know as to where the
2245 next waiter record is;
2246
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002247 (2) read the pointer to the waiter's task structure;
David Howells108b42b2006-03-31 16:00:29 +01002248
2249 (3) clear the task pointer to tell the waiter it has been given the semaphore;
2250
2251 (4) call wake_up_process() on the task; and
2252
2253 (5) release the reference held on the waiter's task struct.
2254
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002255In other words, it has to perform this sequence of events:
David Howells108b42b2006-03-31 16:00:29 +01002256
2257 LOAD waiter->list.next;
2258 LOAD waiter->task;
2259 STORE waiter->task;
2260 CALL wakeup
2261 RELEASE task
2262
2263and if any of these steps occur out of order, then the whole thing may
2264malfunction.
2265
2266Once it has queued itself and dropped the semaphore lock, the waiter does not
2267get the lock again; it instead just waits for its task pointer to be cleared
2268before proceeding. Since the record is on the waiter's stack, this means that
2269if the task pointer is cleared _before_ the next pointer in the list is read,
2270another CPU might start processing the waiter and might clobber the waiter's
2271stack before the up*() function has a chance to read the next pointer.
2272
2273Consider then what might happen to the above sequence of events:
2274
2275 CPU 1 CPU 2
2276 =============================== ===============================
2277 down_xxx()
2278 Queue waiter
2279 Sleep
2280 up_yyy()
2281 LOAD waiter->task;
2282 STORE waiter->task;
2283 Woken up by other event
2284 <preempt>
2285 Resume processing
2286 down_xxx() returns
2287 call foo()
2288 foo() clobbers *waiter
2289 </preempt>
2290 LOAD waiter->list.next;
2291 --- OOPS ---
2292
2293This could be dealt with using the semaphore lock, but then the down_xxx()
2294function has to needlessly get the spinlock again after being woken up.
2295
2296The way to deal with this is to insert a general SMP memory barrier:
2297
2298 LOAD waiter->list.next;
2299 LOAD waiter->task;
2300 smp_mb();
2301 STORE waiter->task;
2302 CALL wakeup
2303 RELEASE task
2304
2305In this case, the barrier makes a guarantee that all memory accesses before the
2306barrier will appear to happen before all the memory accesses after the barrier
2307with respect to the other CPUs on the system. It does _not_ guarantee that all
2308the memory accesses before the barrier will be complete by the time the barrier
2309instruction itself is complete.
2310
2311On a UP system - where this wouldn't be a problem - the smp_mb() is just a
2312compiler barrier, thus making sure the compiler emits the instructions in the
David Howells6bc39272006-06-25 05:49:22 -07002313right order without actually intervening in the CPU. Since there's only one
2314CPU, that CPU's dependency ordering logic will take care of everything else.
David Howells108b42b2006-03-31 16:00:29 +01002315
2316
2317ATOMIC OPERATIONS
2318-----------------
2319
David Howellsdbc87002006-04-10 22:54:23 -07002320Whilst they are technically interprocessor interaction considerations, atomic
2321operations are noted specially as some of them imply full memory barriers and
2322some don't, but they're very heavily relied on as a group throughout the
2323kernel.
2324
2325Any atomic operation that modifies some state in memory and returns information
2326about the state (old or new) implies an SMP-conditional general memory barrier
Nick Piggin26333572007-10-18 03:06:39 -07002327(smp_mb()) on each side of the actual operation (with the exception of
2328explicit lock operations, described later). These include:
David Howells108b42b2006-03-31 16:00:29 +01002329
2330 xchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002331 atomic_xchg(); atomic_long_xchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002332 atomic_inc_return(); atomic_long_inc_return();
2333 atomic_dec_return(); atomic_long_dec_return();
2334 atomic_add_return(); atomic_long_add_return();
2335 atomic_sub_return(); atomic_long_sub_return();
2336 atomic_inc_and_test(); atomic_long_inc_and_test();
2337 atomic_dec_and_test(); atomic_long_dec_and_test();
2338 atomic_sub_and_test(); atomic_long_sub_and_test();
2339 atomic_add_negative(); atomic_long_add_negative();
David Howellsdbc87002006-04-10 22:54:23 -07002340 test_and_set_bit();
2341 test_and_clear_bit();
2342 test_and_change_bit();
David Howells108b42b2006-03-31 16:00:29 +01002343
Will Deaconed2de9f2015-07-16 16:10:06 +01002344 /* when succeeds */
2345 cmpxchg();
2346 atomic_cmpxchg(); atomic_long_cmpxchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002347 atomic_add_unless(); atomic_long_add_unless();
2348
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002349These are used for such things as implementing ACQUIRE-class and RELEASE-class
David Howellsdbc87002006-04-10 22:54:23 -07002350operations and adjusting reference counters towards object destruction, and as
2351such the implicit memory barrier effects are necessary.
David Howells108b42b2006-03-31 16:00:29 +01002352
David Howells108b42b2006-03-31 16:00:29 +01002353
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002354The following operations are potential problems as they do _not_ imply memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002355barriers, but might be used for implementing such things as RELEASE-class
David Howellsdbc87002006-04-10 22:54:23 -07002356operations:
2357
2358 atomic_set();
David Howells108b42b2006-03-31 16:00:29 +01002359 set_bit();
2360 clear_bit();
2361 change_bit();
David Howellsdbc87002006-04-10 22:54:23 -07002362
2363With these the appropriate explicit memory barrier should be used if necessary
Peter Zijlstra1b156112014-03-13 19:00:35 +01002364(smp_mb__before_atomic() for instance).
David Howells108b42b2006-03-31 16:00:29 +01002365
2366
David Howellsdbc87002006-04-10 22:54:23 -07002367The following also do _not_ imply memory barriers, and so may require explicit
Peter Zijlstra1b156112014-03-13 19:00:35 +01002368memory barriers under some circumstances (smp_mb__before_atomic() for
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002369instance):
David Howells108b42b2006-03-31 16:00:29 +01002370
2371 atomic_add();
2372 atomic_sub();
2373 atomic_inc();
2374 atomic_dec();
2375
2376If they're used for statistics generation, then they probably don't need memory
2377barriers, unless there's a coupling between statistical data.
2378
2379If they're used for reference counting on an object to control its lifetime,
2380they probably don't need memory barriers because either the reference count
2381will be adjusted inside a locked section, or the caller will already hold
2382sufficient references to make the lock, and thus a memory barrier unnecessary.
2383
2384If they're used for constructing a lock of some description, then they probably
2385do need memory barriers as a lock primitive generally has to do things in a
2386specific order.
2387
David Howells108b42b2006-03-31 16:00:29 +01002388Basically, each usage case has to be carefully considered as to whether memory
David Howellsdbc87002006-04-10 22:54:23 -07002389barriers are needed or not.
2390
Nick Piggin26333572007-10-18 03:06:39 -07002391The following operations are special locking primitives:
2392
2393 test_and_set_bit_lock();
2394 clear_bit_unlock();
2395 __clear_bit_unlock();
2396
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002397These implement ACQUIRE-class and RELEASE-class operations. These should be used in
Nick Piggin26333572007-10-18 03:06:39 -07002398preference to other operations when implementing locking primitives, because
2399their implementations can be optimised on many architectures.
2400
David Howellsdbc87002006-04-10 22:54:23 -07002401[!] Note that special memory barrier primitives are available for these
2402situations because on some CPUs the atomic instructions used imply full memory
2403barriers, and so barrier instructions are superfluous in conjunction with them,
2404and in such cases the special barrier primitives will be no-ops.
David Howells108b42b2006-03-31 16:00:29 +01002405
2406See Documentation/atomic_ops.txt for more information.
2407
2408
2409ACCESSING DEVICES
2410-----------------
2411
2412Many devices can be memory mapped, and so appear to the CPU as if they're just
2413a set of memory locations. To control such a device, the driver usually has to
2414make the right memory accesses in exactly the right order.
2415
2416However, having a clever CPU or a clever compiler creates a potential problem
2417in that the carefully sequenced accesses in the driver code won't reach the
2418device in the requisite order if the CPU or the compiler thinks it is more
2419efficient to reorder, combine or merge accesses - something that would cause
2420the device to malfunction.
2421
2422Inside of the Linux kernel, I/O should be done through the appropriate accessor
2423routines - such as inb() or writel() - which know how to make such accesses
2424appropriately sequential. Whilst this, for the most part, renders the explicit
2425use of memory barriers unnecessary, there are a couple of situations where they
2426might be needed:
2427
2428 (1) On some systems, I/O stores are not strongly ordered across all CPUs, and
2429 so for _all_ general drivers locks should be used and mmiowb() must be
2430 issued prior to unlocking the critical section.
2431
2432 (2) If the accessor functions are used to refer to an I/O memory window with
2433 relaxed memory access properties, then _mandatory_ memory barriers are
2434 required to enforce ordering.
2435
2436See Documentation/DocBook/deviceiobook.tmpl for more information.
2437
2438
2439INTERRUPTS
2440----------
2441
2442A driver may be interrupted by its own interrupt service routine, and thus the
2443two parts of the driver may interfere with each other's attempts to control or
2444access the device.
2445
2446This may be alleviated - at least in part - by disabling local interrupts (a
2447form of locking), such that the critical operations are all contained within
2448the interrupt-disabled section in the driver. Whilst the driver's interrupt
2449routine is executing, the driver's core may not run on the same CPU, and its
2450interrupt is not permitted to happen again until the current interrupt has been
2451handled, thus the interrupt handler does not need to lock against that.
2452
2453However, consider a driver that was talking to an ethernet card that sports an
2454address register and a data register. If that driver's core talks to the card
2455under interrupt-disablement and then the driver's interrupt handler is invoked:
2456
2457 LOCAL IRQ DISABLE
2458 writew(ADDR, 3);
2459 writew(DATA, y);
2460 LOCAL IRQ ENABLE
2461 <interrupt>
2462 writew(ADDR, 4);
2463 q = readw(DATA);
2464 </interrupt>
2465
2466The store to the data register might happen after the second store to the
2467address register if ordering rules are sufficiently relaxed:
2468
2469 STORE *ADDR = 3, STORE *ADDR = 4, STORE *DATA = y, q = LOAD *DATA
2470
2471
2472If ordering rules are relaxed, it must be assumed that accesses done inside an
2473interrupt disabled section may leak outside of it and may interleave with
2474accesses performed in an interrupt - and vice versa - unless implicit or
2475explicit barriers are used.
2476
2477Normally this won't be a problem because the I/O accesses done inside such
2478sections will include synchronous load operations on strictly ordered I/O
2479registers that form implicit I/O barriers. If this isn't sufficient then an
2480mmiowb() may need to be used explicitly.
2481
2482
2483A similar situation may occur between an interrupt routine and two routines
2484running on separate CPUs that communicate with each other. If such a case is
2485likely, then interrupt-disabling locks should be used to guarantee ordering.
2486
2487
2488==========================
2489KERNEL I/O BARRIER EFFECTS
2490==========================
2491
2492When accessing I/O memory, drivers should use the appropriate accessor
2493functions:
2494
2495 (*) inX(), outX():
2496
2497 These are intended to talk to I/O space rather than memory space, but
2498 that's primarily a CPU-specific concept. The i386 and x86_64 processors do
2499 indeed have special I/O space access cycles and instructions, but many
2500 CPUs don't have such a concept.
2501
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002502 The PCI bus, amongst others, defines an I/O space concept which - on such
2503 CPUs as i386 and x86_64 - readily maps to the CPU's concept of I/O
David Howells6bc39272006-06-25 05:49:22 -07002504 space. However, it may also be mapped as a virtual I/O space in the CPU's
2505 memory map, particularly on those CPUs that don't support alternate I/O
2506 spaces.
David Howells108b42b2006-03-31 16:00:29 +01002507
2508 Accesses to this space may be fully synchronous (as on i386), but
2509 intermediary bridges (such as the PCI host bridge) may not fully honour
2510 that.
2511
2512 They are guaranteed to be fully ordered with respect to each other.
2513
2514 They are not guaranteed to be fully ordered with respect to other types of
2515 memory and I/O operation.
2516
2517 (*) readX(), writeX():
2518
2519 Whether these are guaranteed to be fully ordered and uncombined with
2520 respect to each other on the issuing CPU depends on the characteristics
2521 defined for the memory window through which they're accessing. On later
2522 i386 architecture machines, for example, this is controlled by way of the
2523 MTRR registers.
2524
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002525 Ordinarily, these will be guaranteed to be fully ordered and uncombined,
David Howells108b42b2006-03-31 16:00:29 +01002526 provided they're not accessing a prefetchable device.
2527
2528 However, intermediary hardware (such as a PCI bridge) may indulge in
2529 deferral if it so wishes; to flush a store, a load from the same location
2530 is preferred[*], but a load from the same device or from configuration
2531 space should suffice for PCI.
2532
2533 [*] NOTE! attempting to load from the same location as was written to may
Ingo Molnare0edc782013-11-22 11:24:53 +01002534 cause a malfunction - consider the 16550 Rx/Tx serial registers for
2535 example.
David Howells108b42b2006-03-31 16:00:29 +01002536
2537 Used with prefetchable I/O memory, an mmiowb() barrier may be required to
2538 force stores to be ordered.
2539
2540 Please refer to the PCI specification for more information on interactions
2541 between PCI transactions.
2542
Will Deacona8e0aea2013-09-04 12:30:08 +01002543 (*) readX_relaxed(), writeX_relaxed()
David Howells108b42b2006-03-31 16:00:29 +01002544
Will Deacona8e0aea2013-09-04 12:30:08 +01002545 These are similar to readX() and writeX(), but provide weaker memory
2546 ordering guarantees. Specifically, they do not guarantee ordering with
2547 respect to normal memory accesses (e.g. DMA buffers) nor do they guarantee
2548 ordering with respect to LOCK or UNLOCK operations. If the latter is
2549 required, an mmiowb() barrier can be used. Note that relaxed accesses to
2550 the same peripheral are guaranteed to be ordered with respect to each
2551 other.
David Howells108b42b2006-03-31 16:00:29 +01002552
2553 (*) ioreadX(), iowriteX()
2554
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002555 These will perform appropriately for the type of access they're actually
David Howells108b42b2006-03-31 16:00:29 +01002556 doing, be it inX()/outX() or readX()/writeX().
2557
2558
2559========================================
2560ASSUMED MINIMUM EXECUTION ORDERING MODEL
2561========================================
2562
2563It has to be assumed that the conceptual CPU is weakly-ordered but that it will
2564maintain the appearance of program causality with respect to itself. Some CPUs
2565(such as i386 or x86_64) are more constrained than others (such as powerpc or
2566frv), and so the most relaxed case (namely DEC Alpha) must be assumed outside
2567of arch-specific code.
2568
2569This means that it must be considered that the CPU will execute its instruction
2570stream in any order it feels like - or even in parallel - provided that if an
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002571instruction in the stream depends on an earlier instruction, then that
David Howells108b42b2006-03-31 16:00:29 +01002572earlier instruction must be sufficiently complete[*] before the later
2573instruction may proceed; in other words: provided that the appearance of
2574causality is maintained.
2575
2576 [*] Some instructions have more than one effect - such as changing the
2577 condition codes, changing registers or changing memory - and different
2578 instructions may depend on different effects.
2579
2580A CPU may also discard any instruction sequence that winds up having no
2581ultimate effect. For example, if two adjacent instructions both load an
2582immediate value into the same register, the first may be discarded.
2583
2584
2585Similarly, it has to be assumed that compiler might reorder the instruction
2586stream in any way it sees fit, again provided the appearance of causality is
2587maintained.
2588
2589
2590============================
2591THE EFFECTS OF THE CPU CACHE
2592============================
2593
2594The way cached memory operations are perceived across the system is affected to
2595a certain extent by the caches that lie between CPUs and memory, and by the
2596memory coherence system that maintains the consistency of state in the system.
2597
2598As far as the way a CPU interacts with another part of the system through the
2599caches goes, the memory system has to include the CPU's caches, and memory
2600barriers for the most part act at the interface between the CPU and its cache
2601(memory barriers logically act on the dotted line in the following diagram):
2602
2603 <--- CPU ---> : <----------- Memory ----------->
2604 :
2605 +--------+ +--------+ : +--------+ +-----------+
2606 | | | | : | | | | +--------+
Ingo Molnare0edc782013-11-22 11:24:53 +01002607 | CPU | | Memory | : | CPU | | | | |
2608 | Core |--->| Access |----->| Cache |<-->| | | |
David Howells108b42b2006-03-31 16:00:29 +01002609 | | | Queue | : | | | |--->| Memory |
Ingo Molnare0edc782013-11-22 11:24:53 +01002610 | | | | : | | | | | |
2611 +--------+ +--------+ : +--------+ | | | |
David Howells108b42b2006-03-31 16:00:29 +01002612 : | Cache | +--------+
2613 : | Coherency |
2614 : | Mechanism | +--------+
2615 +--------+ +--------+ : +--------+ | | | |
2616 | | | | : | | | | | |
2617 | CPU | | Memory | : | CPU | | |--->| Device |
Ingo Molnare0edc782013-11-22 11:24:53 +01002618 | Core |--->| Access |----->| Cache |<-->| | | |
2619 | | | Queue | : | | | | | |
David Howells108b42b2006-03-31 16:00:29 +01002620 | | | | : | | | | +--------+
2621 +--------+ +--------+ : +--------+ +-----------+
2622 :
2623 :
2624
2625Although any particular load or store may not actually appear outside of the
2626CPU that issued it since it may have been satisfied within the CPU's own cache,
2627it will still appear as if the full memory access had taken place as far as the
2628other CPUs are concerned since the cache coherency mechanisms will migrate the
2629cacheline over to the accessing CPU and propagate the effects upon conflict.
2630
2631The CPU core may execute instructions in any order it deems fit, provided the
2632expected program causality appears to be maintained. Some of the instructions
2633generate load and store operations which then go into the queue of memory
2634accesses to be performed. The core may place these in the queue in any order
2635it wishes, and continue execution until it is forced to wait for an instruction
2636to complete.
2637
2638What memory barriers are concerned with is controlling the order in which
2639accesses cross from the CPU side of things to the memory side of things, and
2640the order in which the effects are perceived to happen by the other observers
2641in the system.
2642
2643[!] Memory barriers are _not_ needed within a given CPU, as CPUs always see
2644their own loads and stores as if they had happened in program order.
2645
2646[!] MMIO or other device accesses may bypass the cache system. This depends on
2647the properties of the memory window through which devices are accessed and/or
2648the use of any special device communication instructions the CPU may have.
2649
2650
2651CACHE COHERENCY
2652---------------
2653
2654Life isn't quite as simple as it may appear above, however: for while the
2655caches are expected to be coherent, there's no guarantee that that coherency
2656will be ordered. This means that whilst changes made on one CPU will
2657eventually become visible on all CPUs, there's no guarantee that they will
2658become apparent in the same order on those other CPUs.
2659
2660
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002661Consider dealing with a system that has a pair of CPUs (1 & 2), each of which
2662has a pair of parallel data caches (CPU 1 has A/B, and CPU 2 has C/D):
David Howells108b42b2006-03-31 16:00:29 +01002663
2664 :
2665 : +--------+
2666 : +---------+ | |
2667 +--------+ : +--->| Cache A |<------->| |
2668 | | : | +---------+ | |
2669 | CPU 1 |<---+ | |
2670 | | : | +---------+ | |
2671 +--------+ : +--->| Cache B |<------->| |
2672 : +---------+ | |
2673 : | Memory |
2674 : +---------+ | System |
2675 +--------+ : +--->| Cache C |<------->| |
2676 | | : | +---------+ | |
2677 | CPU 2 |<---+ | |
2678 | | : | +---------+ | |
2679 +--------+ : +--->| Cache D |<------->| |
2680 : +---------+ | |
2681 : +--------+
2682 :
2683
2684Imagine the system has the following properties:
2685
2686 (*) an odd-numbered cache line may be in cache A, cache C or it may still be
2687 resident in memory;
2688
2689 (*) an even-numbered cache line may be in cache B, cache D or it may still be
2690 resident in memory;
2691
2692 (*) whilst the CPU core is interrogating one cache, the other cache may be
2693 making use of the bus to access the rest of the system - perhaps to
2694 displace a dirty cacheline or to do a speculative load;
2695
2696 (*) each cache has a queue of operations that need to be applied to that cache
2697 to maintain coherency with the rest of the system;
2698
2699 (*) the coherency queue is not flushed by normal loads to lines already
2700 present in the cache, even though the contents of the queue may
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002701 potentially affect those loads.
David Howells108b42b2006-03-31 16:00:29 +01002702
2703Imagine, then, that two writes are made on the first CPU, with a write barrier
2704between them to guarantee that they will appear to reach that CPU's caches in
2705the requisite order:
2706
2707 CPU 1 CPU 2 COMMENT
2708 =============== =============== =======================================
2709 u == 0, v == 1 and p == &u, q == &u
2710 v = 2;
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002711 smp_wmb(); Make sure change to v is visible before
David Howells108b42b2006-03-31 16:00:29 +01002712 change to p
2713 <A:modify v=2> v is now in cache A exclusively
2714 p = &v;
2715 <B:modify p=&v> p is now in cache B exclusively
2716
2717The write memory barrier forces the other CPUs in the system to perceive that
2718the local CPU's caches have apparently been updated in the correct order. But
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002719now imagine that the second CPU wants to read those values:
David Howells108b42b2006-03-31 16:00:29 +01002720
2721 CPU 1 CPU 2 COMMENT
2722 =============== =============== =======================================
2723 ...
2724 q = p;
2725 x = *q;
2726
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002727The above pair of reads may then fail to happen in the expected order, as the
David Howells108b42b2006-03-31 16:00:29 +01002728cacheline holding p may get updated in one of the second CPU's caches whilst
2729the update to the cacheline holding v is delayed in the other of the second
2730CPU's caches by some other cache event:
2731
2732 CPU 1 CPU 2 COMMENT
2733 =============== =============== =======================================
2734 u == 0, v == 1 and p == &u, q == &u
2735 v = 2;
2736 smp_wmb();
2737 <A:modify v=2> <C:busy>
2738 <C:queue v=2>
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002739 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002740 <D:request p>
2741 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002742 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002743 x = *q;
2744 <C:read *q> Reads from v before v updated in cache
2745 <C:unbusy>
2746 <C:commit v=2>
2747
2748Basically, whilst both cachelines will be updated on CPU 2 eventually, there's
2749no guarantee that, without intervention, the order of update will be the same
2750as that committed on CPU 1.
2751
2752
2753To intervene, we need to interpolate a data dependency barrier or a read
2754barrier between the loads. This will force the cache to commit its coherency
2755queue before processing any further requests:
2756
2757 CPU 1 CPU 2 COMMENT
2758 =============== =============== =======================================
2759 u == 0, v == 1 and p == &u, q == &u
2760 v = 2;
2761 smp_wmb();
2762 <A:modify v=2> <C:busy>
2763 <C:queue v=2>
Paolo 'Blaisorblade' Giarrusso3fda9822006-10-19 23:28:19 -07002764 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002765 <D:request p>
2766 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002767 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002768 smp_read_barrier_depends()
2769 <C:unbusy>
2770 <C:commit v=2>
2771 x = *q;
2772 <C:read *q> Reads from v after v updated in cache
2773
2774
2775This sort of problem can be encountered on DEC Alpha processors as they have a
2776split cache that improves performance by making better use of the data bus.
2777Whilst most CPUs do imply a data dependency barrier on the read when a memory
2778access depends on a read, not all do, so it may not be relied on.
2779
2780Other CPUs may also have split caches, but must coordinate between the various
Matt LaPlante3f6dee92006-10-03 22:45:33 +02002781cachelets for normal memory accesses. The semantics of the Alpha removes the
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002782need for coordination in the absence of memory barriers.
David Howells108b42b2006-03-31 16:00:29 +01002783
2784
2785CACHE COHERENCY VS DMA
2786----------------------
2787
2788Not all systems maintain cache coherency with respect to devices doing DMA. In
2789such cases, a device attempting DMA may obtain stale data from RAM because
2790dirty cache lines may be resident in the caches of various CPUs, and may not
2791have been written back to RAM yet. To deal with this, the appropriate part of
2792the kernel must flush the overlapping bits of cache on each CPU (and maybe
2793invalidate them as well).
2794
2795In addition, the data DMA'd to RAM by a device may be overwritten by dirty
2796cache lines being written back to RAM from a CPU's cache after the device has
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002797installed its own data, or cache lines present in the CPU's cache may simply
2798obscure the fact that RAM has been updated, until at such time as the cacheline
2799is discarded from the CPU's cache and reloaded. To deal with this, the
2800appropriate part of the kernel must invalidate the overlapping bits of the
David Howells108b42b2006-03-31 16:00:29 +01002801cache on each CPU.
2802
2803See Documentation/cachetlb.txt for more information on cache management.
2804
2805
2806CACHE COHERENCY VS MMIO
2807-----------------------
2808
2809Memory mapped I/O usually takes place through memory locations that are part of
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002810a window in the CPU's memory space that has different properties assigned than
David Howells108b42b2006-03-31 16:00:29 +01002811the usual RAM directed window.
2812
2813Amongst these properties is usually the fact that such accesses bypass the
2814caching entirely and go directly to the device buses. This means MMIO accesses
2815may, in effect, overtake accesses to cached memory that were emitted earlier.
2816A memory barrier isn't sufficient in such a case, but rather the cache must be
2817flushed between the cached memory write and the MMIO access if the two are in
2818any way dependent.
2819
2820
2821=========================
2822THE THINGS CPUS GET UP TO
2823=========================
2824
2825A programmer might take it for granted that the CPU will perform memory
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002826operations in exactly the order specified, so that if the CPU is, for example,
David Howells108b42b2006-03-31 16:00:29 +01002827given the following piece of code to execute:
2828
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002829 a = READ_ONCE(*A);
2830 WRITE_ONCE(*B, b);
2831 c = READ_ONCE(*C);
2832 d = READ_ONCE(*D);
2833 WRITE_ONCE(*E, e);
David Howells108b42b2006-03-31 16:00:29 +01002834
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002835they would then expect that the CPU will complete the memory operation for each
David Howells108b42b2006-03-31 16:00:29 +01002836instruction before moving on to the next one, leading to a definite sequence of
2837operations as seen by external observers in the system:
2838
2839 LOAD *A, STORE *B, LOAD *C, LOAD *D, STORE *E.
2840
2841
2842Reality is, of course, much messier. With many CPUs and compilers, the above
2843assumption doesn't hold because:
2844
2845 (*) loads are more likely to need to be completed immediately to permit
2846 execution progress, whereas stores can often be deferred without a
2847 problem;
2848
2849 (*) loads may be done speculatively, and the result discarded should it prove
2850 to have been unnecessary;
2851
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002852 (*) loads may be done speculatively, leading to the result having been fetched
2853 at the wrong time in the expected sequence of events;
David Howells108b42b2006-03-31 16:00:29 +01002854
2855 (*) the order of the memory accesses may be rearranged to promote better use
2856 of the CPU buses and caches;
2857
2858 (*) loads and stores may be combined to improve performance when talking to
2859 memory or I/O hardware that can do batched accesses of adjacent locations,
2860 thus cutting down on transaction setup costs (memory and PCI devices may
2861 both be able to do this); and
2862
2863 (*) the CPU's data cache may affect the ordering, and whilst cache-coherency
2864 mechanisms may alleviate this - once the store has actually hit the cache
2865 - there's no guarantee that the coherency management will be propagated in
2866 order to other CPUs.
2867
2868So what another CPU, say, might actually observe from the above piece of code
2869is:
2870
2871 LOAD *A, ..., LOAD {*C,*D}, STORE *E, STORE *B
2872
2873 (Where "LOAD {*C,*D}" is a combined load)
2874
2875
2876However, it is guaranteed that a CPU will be self-consistent: it will see its
2877_own_ accesses appear to be correctly ordered, without the need for a memory
2878barrier. For instance with the following code:
2879
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002880 U = READ_ONCE(*A);
2881 WRITE_ONCE(*A, V);
2882 WRITE_ONCE(*A, W);
2883 X = READ_ONCE(*A);
2884 WRITE_ONCE(*A, Y);
2885 Z = READ_ONCE(*A);
David Howells108b42b2006-03-31 16:00:29 +01002886
2887and assuming no intervention by an external influence, it can be assumed that
2888the final result will appear to be:
2889
2890 U == the original value of *A
2891 X == W
2892 Z == Y
2893 *A == Y
2894
2895The code above may cause the CPU to generate the full sequence of memory
2896accesses:
2897
2898 U=LOAD *A, STORE *A=V, STORE *A=W, X=LOAD *A, STORE *A=Y, Z=LOAD *A
2899
2900in that order, but, without intervention, the sequence may have almost any
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002901combination of elements combined or discarded, provided the program's view
2902of the world remains consistent. Note that READ_ONCE() and WRITE_ONCE()
2903are -not- optional in the above example, as there are architectures
2904where a given CPU might reorder successive loads to the same location.
2905On such architectures, READ_ONCE() and WRITE_ONCE() do whatever is
2906necessary to prevent this, for example, on Itanium the volatile casts
2907used by READ_ONCE() and WRITE_ONCE() cause GCC to emit the special ld.acq
2908and st.rel instructions (respectively) that prevent such reordering.
David Howells108b42b2006-03-31 16:00:29 +01002909
2910The compiler may also combine, discard or defer elements of the sequence before
2911the CPU even sees them.
2912
2913For instance:
2914
2915 *A = V;
2916 *A = W;
2917
2918may be reduced to:
2919
2920 *A = W;
2921
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002922since, without either a write barrier or an WRITE_ONCE(), it can be
Paul E. McKenney2ecf8102013-12-11 13:59:04 -08002923assumed that the effect of the storage of V to *A is lost. Similarly:
David Howells108b42b2006-03-31 16:00:29 +01002924
2925 *A = Y;
2926 Z = *A;
2927
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002928may, without a memory barrier or an READ_ONCE() and WRITE_ONCE(), be
2929reduced to:
David Howells108b42b2006-03-31 16:00:29 +01002930
2931 *A = Y;
2932 Z = Y;
2933
2934and the LOAD operation never appear outside of the CPU.
2935
2936
2937AND THEN THERE'S THE ALPHA
2938--------------------------
2939
2940The DEC Alpha CPU is one of the most relaxed CPUs there is. Not only that,
2941some versions of the Alpha CPU have a split data cache, permitting them to have
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002942two semantically-related cache lines updated at separate times. This is where
David Howells108b42b2006-03-31 16:00:29 +01002943the data dependency barrier really becomes necessary as this synchronises both
2944caches with the memory coherence system, thus making it seem like pointer
2945changes vs new data occur in the right order.
2946
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002947The Alpha defines the Linux kernel's memory barrier model.
David Howells108b42b2006-03-31 16:00:29 +01002948
2949See the subsection on "Cache Coherency" above.
2950
2951
David Howells90fddab2010-03-24 09:43:00 +00002952============
2953EXAMPLE USES
2954============
2955
2956CIRCULAR BUFFERS
2957----------------
2958
2959Memory barriers can be used to implement circular buffering without the need
2960of a lock to serialise the producer with the consumer. See:
2961
2962 Documentation/circular-buffers.txt
2963
2964for details.
2965
2966
David Howells108b42b2006-03-31 16:00:29 +01002967==========
2968REFERENCES
2969==========
2970
2971Alpha AXP Architecture Reference Manual, Second Edition (Sites & Witek,
2972Digital Press)
2973 Chapter 5.2: Physical Address Space Characteristics
2974 Chapter 5.4: Caches and Write Buffers
2975 Chapter 5.5: Data Sharing
2976 Chapter 5.6: Read/Write Ordering
2977
2978AMD64 Architecture Programmer's Manual Volume 2: System Programming
2979 Chapter 7.1: Memory-Access Ordering
2980 Chapter 7.4: Buffering and Combining Memory Writes
2981
2982IA-32 Intel Architecture Software Developer's Manual, Volume 3:
2983System Programming Guide
2984 Chapter 7.1: Locked Atomic Operations
2985 Chapter 7.2: Memory Ordering
2986 Chapter 7.4: Serializing Instructions
2987
2988The SPARC Architecture Manual, Version 9
2989 Chapter 8: Memory Models
2990 Appendix D: Formal Specification of the Memory Models
2991 Appendix J: Programming with the Memory Models
2992
2993UltraSPARC Programmer Reference Manual
2994 Chapter 5: Memory Accesses and Cacheability
2995 Chapter 15: Sparc-V9 Memory Models
2996
2997UltraSPARC III Cu User's Manual
2998 Chapter 9: Memory Models
2999
3000UltraSPARC IIIi Processor User's Manual
3001 Chapter 8: Memory Models
3002
3003UltraSPARC Architecture 2005
3004 Chapter 9: Memory
3005 Appendix D: Formal Specifications of the Memory Models
3006
3007UltraSPARC T1 Supplement to the UltraSPARC Architecture 2005
3008 Chapter 8: Memory Models
3009 Appendix F: Caches and Cache Coherency
3010
3011Solaris Internals, Core Kernel Architecture, p63-68:
3012 Chapter 3.3: Hardware Considerations for Locks and
3013 Synchronization
3014
3015Unix Systems for Modern Architectures, Symmetric Multiprocessing and Caching
3016for Kernel Programmers:
3017 Chapter 13: Other Memory Models
3018
3019Intel Itanium Architecture Software Developer's Manual: Volume 1:
3020 Section 2.6: Speculation
3021 Section 4.4: Memory Access