What exact rules in the C++ memory model prevent reordering before acquire operations?

问题

I have a question regarding the order of operations in the following code:

std::atomic<int> x;
std::atomic<int> y;
int r1;
int r2;
void thread1() {
  y.exchange(1, std::memory_order_acq_rel);
  r1 = x.load(std::memory_order_relaxed);
}
void thread2() {
  x.exchange(1, std::memory_order_acq_rel);
  r2 = y.load(std::memory_order_relaxed);
}

Given the description of std::memory_order_acquire on the cppreference page (https://en.cppreference.com/w/cpp/atomic/memory_order), that

A load operation with this memory order performs the acquire operation on the affected memory location: no reads or writes in the current thread can be reordered before this load.

it seems obvious that there can never be an outcome that r1 == 0 && r2 == 0 after running thread1 and thread2 concurrently.

However, I cannot find any wording in the C++ standard (looking at the C++14 draft right now), which establishes guarantees that two relaxed loads cannot be reordered with acquire-release exchanges. What am I missing?

EDIT: As has been suggested in the comments, it is actually possible to get both r1 and r2 equal to zero. I've updated the program to use load-acquire as follows:

std::atomic<int> x;
std::atomic<int> y;
int r1;
int r2;
void thread1() {
  y.exchange(1, std::memory_order_acq_rel);
  r1 = x.load(std::memory_order_acquire);
}
void thread2() {
  x.exchange(1, std::memory_order_acq_rel);
  r2 = y.load(std::memory_order_acquire);
}

Now is it possible to get both and r1 and r2 equal to 0 after concurrently executing thread1 and thread2? If not, which C++ rules prevent this?

回答1:

The standard does not define the C++ memory model in terms of how operations are ordered around atomic operations with a specific ordering parameter. Instead, for the acquire/release ordering model, it defines formal relationships such as "synchronizes-with" and "happens-before" that specify how data is synchronized between threads.

N4762, §29.4.2 - [atomics.order]

An atomic operation A that performs a release operation on an atomic object M synchronizes with an atomic operation B that performs an acquire operation on M and takes its value from any side effect in the release sequence headed by A.

In §6.8.2.1-9, the standard also states that if a store A synchronizes with a load B, anything sequenced before A inter-thread "happens-before" anything sequenced after B.

No "synchronizes-with" (and hence inter-thread happens-before) relationship is established in your second example (the first is even weaker) because the runtime relationships (that check the return values from the loads) are missing.
But even if you did check the return value, it would not be helpful since the exchange operations do not actually 'release' anything (i.e. no memory operations are sequenced before those operations). Neiter do the atomic load operations 'acquire' anything since no operations are sequenced after the loads.

Therefore, according to the standard, each of the four possible outcomes for the loads in both examples (including 0 0) is valid. In fact, the guarantees given by the standard are no stronger than memory_order_relaxed on all operations.

If you want to exclude the 0 0 result in your code, all 4 operations must use std::memory_order_seq_cst. That guarantees a single total order of the involved operations.

回答2:

in Release-Acquire ordering for create synchronization point between 2 threads we need some atomic object M which will be the same in both operations

An atomic operation A that performs a release operation on an atomic object M synchronizes with an atomic operation B that performs an acquire operation on M and takes its value from any side effect in the release sequence headed by A.

or in more details:

If an atomic store in thread A is tagged memory_order_release and an atomic load in thread B from the same variable is tagged memory_order_acquire, all memory writes (non-atomic and relaxed atomic) that happened-before the atomic store from the point of view of thread A, become visible side-effects in thread B. That is, once the atomic load is completed, thread B is guaranteed to see everything thread A wrote to memory.

The synchronization is established only between the threads releasing and acquiring the same atomic variable.

     N = u                |  if (M.load(acquire) == v)    :[B]
[A]: M.store(v, release)  |  assert(N == u)

here synchronization point on M store-release and load-acquire(which take value from store-release !). as result store N = u in thread A (before store-release on M) visible in B (N == u) after load-acquire on same M

if take example:

atomic<int> x, y;
int r1, r2;

void thread_A() {
  y.exchange(1, memory_order_acq_rel);
  r1 = x.load(memory_order_acquire);
}
void thread_B() {
  x.exchange(1, memory_order_acq_rel);
  r2 = y.load(memory_order_acquire);
}

what we can select for common atomic object M ? say x ? x.load(memory_order_acquire); will be synchronization point with x.exchange(1, memory_order_acq_rel) ( memory_order_acq_rel include memory_order_release (more strong) and exchange include store) if x.load load value from x.exchange and main will be synchronized loads after acquire (be in code after acquire nothing exist) with stores before release (but again before exchange nothing in code).

correct solution (look for almost exactly question ) can be next:

atomic<int> x, y;
int r1, r2;

void thread_A()
{
    x.exchange(1, memory_order_acq_rel); // [Ax]
    r1 = y.exchange(1, memory_order_acq_rel); // [Ay]
}

void thread_B()
{
    y.exchange(1, memory_order_acq_rel); // [By]
    r2 = x.exchange(1, memory_order_acq_rel); // [Bx]
}

assume that r1 == 0.

All modifications to any particular atomic variable occur in a total order that is specific to this one atomic variable.

we have 2 modification of y : [Ay] and [By]. because r1 == 0 this mean that [Ay] happens before [By] in total modification order of y. from this - [By] read value stored by [Ay]. so we have next:

A is write to x - [Ax]
A do store-release [Ay] to y after this ( acq_rel include release, exchange include store)
B load-acquire from y ([By] value stored by [Ay]
once the atomic load-acquire (on y) is completed, thread B is guaranteed to see everything thread A wrote to memory before store-release (on y). so it view side-effect of [Ax] - and r2 == 1

another possible solution use atomic_thread_fence

atomic<int> x, y;
int r1, r2;

void thread_A()
{
    x.store(1, memory_order_relaxed); // [A1]
    atomic_thread_fence(memory_order_acq_rel); // [A2]
    r1 = y.exchange(1, memory_order_relaxed); // [A3]
}

void thread_B()
{
    y.store(1, memory_order_relaxed); // [B1]
    atomic_thread_fence(memory_order_acq_rel); // [B2]
    r2 = x.exchange(1, memory_order_relaxed); // [B3]
}

again because all modifications of atomic variable y occur in a total order. [A3] will be before [B1] or visa versa.

if [B1] before [A3] - [A3] read value stored by [B1] => r1 == 1.
if [A3] before [B1] - the [B1] is read value stored by [A3] and from Fence-fence synchronization:

A release fence [A2] in thread A synchronizes-with an acquire fence [B2] in thread B, if:

There exists an atomic object y,
There exists an atomic write [A3] (with any memory order) that modifies y in thread A
[A2] is sequenced-before [A3] in thread A
There exists an atomic read [B1] (with any memory order) in thread B
[B1] reads the value written by [A3]
[B1] is sequenced-before [B2] in thread B

In this case, all stores ([A1]) that are sequenced-before [A2] in thread A will happen-before all loads ([B3]) from the same locations (x) made in thread B after [B2]

so [A1] (store 1 to x) will be before and have visible effect for [B3] (load form x and save result to r2 ). so will be loaded 1 from x and r2==1

[A1]: x = 1               |  if (y.load(relaxed) == 1) :[B1]
[A2]: ### release ###     |  ### acquire ###           :[B2]
[A3]: y.store(1, relaxed) |  assert(x == 1)            :[B3]

回答3:

You already have an answer to the language-lawyer part of this. But I want to answer the related question of how to understand why this can be possible in asm on a possible CPU architecture that uses LL/SC for RMW atomics.

It doesn't make sense for C++11 to forbid this reordering: it would require a store-load barrier in this case where some CPU architectures could avoid one.

It might actually be possible with real compilers on PowerPC, given the way they map C++11 memory-orders to asm instructions.

On PowerPC64, a function with an acq_rel exchange and an acquire load (using pointer args instead of static variables) compiles as follows with gcc6.3 -O3 -mregnames. This is from a C11 version because I wanted to look at clang output for MIPS and SPARC, and Godbolt's clang setup works for C11 <atomic.h> but fails for C++11 <atomic> when you use -target sparc64.

(source + asm on Godbolt for MIPS32R6, SPARC64, ARM 32, and PowerPC64.)

foo:
    lwsync            # with seq_cst exchange this is full sync, not just lwsync
                      # gone if we use exchage with mo_acquire or relaxed
                      # so this barrier is providing release-store ordering
    li %r9,1
.L2:
    lwarx %r10,0,%r4    # load-linked from 0(%r4)
    stwcx. %r9,0,%r4    # store-conditional 0(%r4)
    bne %cr0,.L2        # retry if SC failed
    isync             # missing if we use exchange(1, mo_release) or relaxed

    ld %r3,0(%r3)       # 64-bit load double-word of *a
    cmpw %cr7,%r3,%r3
    bne- %cr7,$+4       # skip over the isync if something about the load? PowerPC is weird
    isync             # make the *a load a load-acquire
    blr

isync is not a store-load barrier; it only requires the preceding instructions to complete locally (retire from the out-of-order part of the core). It doesn't wait for the store buffer to be flushed so other threads can see the earlier stores.

Thus the SC (stwcx.) store that's part of the exchange can sit in the store buffer and become globally visible after the pure acquire-load that follows it. In fact, another Q&A already asked this, and the answer is that we think this reordering is possible. Does `isync` prevent Store-Load reordering on CPU PowerPC?

If the pure load is seq_cst, PowerPC64 gcc puts a sync before the ld. Making the exchange seq_cst does not prevent the reordering. Remember that C++11 only guarantees a single total order for SC operations, so the exchange and the load both need to be SC for C++11 to guarantee it.

So PowerPC has a bit of an unusual mapping from C++11 to asm for atomics. Most systems put the heavier barriers on stores, allowing seq-cst loads to be cheaper or only have a barrier on one side. I'm not sure if this was required for PowerPC's famously-weak memory ordering, or if another choice was possible.

https://www.cl.cam.ac.uk/~pes20/cpp/cpp0xmappings.html shows some possible implementations on various architectures. It mentions multiple alternatives for ARM.

On AArch64, we get this for the original C++ version of thread1:

thread1():
    adrp    x0, .LANCHOR0
    mov     w1, 1
    add     x0, x0, :lo12:.LANCHOR0
.L2:
    ldaxr   w2, [x0]            @ load-linked with acquire semantics
    stlxr   w3, w1, [x0]        @ store-conditional with sc-release semantics
    cbnz    w3, .L2             @ retry until exchange succeeds

    add     x1, x0, 8           @ the compiler noticed the variables were next to each other
    ldar    w1, [x1]            @ load-acquire

    str     w1, [x0, 12]        @ r1 = load result
    ret

The reordering can't happen there because AArch64 release-stores are sequential-release, not plain release. This means they can't reorder with later loads.

But on a hypothetical machine that also or instead had plain-release LL/SC atomics, it's easy to see that an acq_rel doesn't stop later loads to different cache lines from becoming globally visible after the LL but before the SC of the exchange.

If exchange is implemented with a single transaction like on x86, so the load and store are adjacent in the global order of memory operations, then certainly no later operations can be reordered with an acq_rel exchange and it's basically equivalent to seq_cst.

But LL/SC doesn't have to be a true atomic transaction to give RMW atomicity for that location.

In fact, a single asm swap instruction could have relaxed or acq_rel semantics. SPARC64 needs membar instructions around its swap instruction, so unlike x86's xchg it's not seq-cst on its own. (SPARC has really nice / human readable instruction mnemonics, especially compared to PowerPC. Well basically anything is more readable that PowerPC.)

Thus it doesn't make sense for C++11 to require that it did: it would hurt an implementation on a CPU that didn't otherwise need a store-load barrier.

回答4:

As language lawyer reasonings are hard to follow, I thought I'd add how a programmer who understands atomics would reason about the second snippet in your question:

Since this is symmetrical code, it is enough to look at just one side. Since the question is about the value of r1 (r2), we start with looking at

r1 = x.load(std::memory_order_acquire);

Depending on what the value of r1 is, we can say something about the visibility of other values. However, since the value of r1 isn't tested - the acquire is irrelevant. In either case, the value of r1 can be any value that was ever written to it (in the past or future *)). Therefore it can be zero. Nevertheless, we can assume it to BE zero because we're interested in whether or not the outcome of the whole program can be 0 0, which is a sort of testing the value of r1.

Hence, assuming we read zero THEN we can say that if that zero was written by another thread with memory_order_release then every other write to memory done by that thread before the store release will also be visible to this thread. However, value of zero that we read is the initialization value of x, and initialization values are non-atomic - let alone a 'release' - and certainly there wasn't anything "ordered" in front of them in terms of writing that value to memory; so there is nothing we can say about the visibility of other memory locations. In other words, again, the 'acquire' is irrelevant.

So, we can get r1 = 0 and the fact that we used acquire is irrelevant. The same reasoning then holds of r2. So the result can be r1 = r2 = 0.

In fact, if you assume the value of r1 is 1 after the load acquire, and that that 1 was written by thread2 with memory order release (which MUST be the case, since is the only place where a value of 1 is ever written to x) then all we know is that everything written to memory by thread2 before that store release will also be visible to thread1 (provided thread1 read x == 1 thus!). But thread2 doesn't write ANYTHING before writing to x, so again the whole release-acquire relationship is irrelevant, even in the case of loading a value of 1.

*) However, it is possible with further reasoning to show that certain value can never occur because of inconsistency with the memory model - but that doesn't happen here.

回答5:

In the original version, it is possible to see r1 == 0 && r2 == 0 because there is no requirement that the stores propogate to the other thread before it reads it. This is not a re-ordering of either thread's operations, but e.g. a read of stale cache.

Thread 1's cache   |   Thread 2's cache
  x == 0;          |     x == 0;
  y == 0;          |     y == 0;

y.exchange(1, std::memory_order_acq_rel); // Thread 1
x.exchange(1, std::memory_order_acq_rel); // Thread 2

The release on Thread 1 is ignored by Thread 2, and vice-versa. In the abstract machine there is not consistency with the values of x and y on the threads

Thread 1's cache   |   Thread 2's cache
  x == 0; // stale |     x == 1;
  y == 1;          |     y == 0; // stale

r1 = x.load(std::memory_order_relaxed); // Thread 1
r2 = y.load(std::memory_order_relaxed); // Thread 2

You need more threads to get "violations of causality" with acquire / release pairs, as the normal ordering rules, combined with the "becomes visible side effect in" rules force at least one of the loads to see 1.

Without loss of generality, let's assume that Thread 1 executes first.

Thread 1's cache   |   Thread 2's cache
  x == 0;          |     x == 0;
  y == 0;          |     y == 0;

y.exchange(1, std::memory_order_acq_rel); // Thread 1

Thread 1's cache   |   Thread 2's cache
  x == 0;          |     x == 0;
  y == 1;          |     y == 1; // sync

The release on Thread 1 forms a pair with the acquire on Thread 2, and the abstract machine describes a consistent y on both threads

r1 = x.load(std::memory_order_relaxed); // Thread 1
x.exchange(1, std::memory_order_acq_rel); // Thread 2
r2 = y.load(std::memory_order_relaxed); // Thread 2

来源：https://stackoverflow.com/questions/52606524/what-exact-rules-in-the-c-memory-model-prevent-reordering-before-acquire-opera

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