C++ 标准：可以将宽松的原子存储提升到互斥锁之上吗？

Question

Is there any wording in the standard that guarantees that relaxed stores to atomics won't be lifted above the locking of a mutex? If not, is there any wording that explicitly says that it's kosher for the compiler or CPU to do so?

For example, take the following program (which could potentially use acq/rel for foo_has_been_set and avoid the lock, and/or make foo itself atomic. It's written this way to illustrate this question.)

std::mutex mu;
int foo = 0;  // Guarded by mu
std::atomic<bool> foo_has_been_set{false};

void SetFoo() {
  mu.lock();
  foo = 1;
  foo_has_been_set.store(true, std::memory_order_relaxed);
  mu.unlock();
}

void CheckFoo() {
  if (foo_has_been_set.load(std::memory_order_relaxed)) {
    mu.lock();
    assert(foo == 1);
    mu.unlock();
  }
}

Is it possible for CheckFoo to crash in the above program if another thread is calling SetFoo concurrently, or is there some guarantee that the store to foo_has_been_set can't be lifted above the call to mu.lock by the compiler and CPU?

This is related to an older question , but it's not 100% clear to me that the answer there applies to this. In particular, the counter-example in that question's answer may apply to two concurrent calls to SetFoo , but I'm interested in the case where the compiler knows that there is one call to SetFoo and one call to CheckFoo . Is that guaranteed to be safe?

I'm looking for specific citations in the standard.

Answer 1

I think I've figured out the particular partial order edges that guarantee the program can't crash. In the answer below I'm referencing version N4659 of the draft standard.

The code involved for the writer thread A and reader thread B is:

A1: mu.lock()
A2: foo = 1
A3: foo_has_been_set.store(relaxed)
A4: mu.unlock()

B1: foo_has_been_set.load(relaxed) <-- (stop if false)
B2: mu.lock()
B3: assert(foo == 1)
B4: mu.unlock()

We seek a proof that if B3 executes, then A2 happens before B3, as defined in [intro.races]/10 . By [intro.races]/10.2 , it's sufficient to prove that A2 inter-thread happens before B3.

Because lock and unlock operations on a given mutex happen in a single total order ( [thread.mutex.requirements.mutex]/5 ), we must have either A1 or B2 coming first. The two cases:

1. Assume that A1 happens before B2. Then by [thread.mutex.class]/1 and [thread.mutex.requirements.mutex]/25 , we know that A4 will synchronize with B2. Therefore by [intro.races]/9.1 , A4 inter-thread happens before B2. Since B2 is sequenced before B3, by [intro.races]/9.3.1 we know that A4 inter-thread happens before B3. Since A2 is sequenced before A4, by [intro.races]/9.3.2 , A2 inter-thread happens before B3.

2. Assume that B2 happens before A1. Then by the same logic as above, we know that B4 synchronizes with A1. So since A1 is sequenced before A3, by [intro.races]/9.3.1 , B4 inter-thread happens before A3. Therefore since B1 is sequenced before B4, by [intro.races]/9.3.2 , B1 inter-thread happens before A3. Therefore by [intro.races]/10.2 , B1 happens before A3. But then according to [intro.races]/16 , B1 must take its value from the pre-A3 state. Therefore the load will return false, and B2 will never run in the first place. In other words, this case can't happen.

So if B3 executes at all (case 1), A2 happens before B3 and the assert will pass. ∎

Answer 2

No memory operation inside a mutex protected region can 'escape' from that area. That applies to all memory operations, atomic and non-atomic.

In section 1.10.1:

a call that acquires a mutex will perform an acquire operation on the locations comprising the mutex Correspondingly, a call that releases the same mutex will perform a release operation on those same locations

Furthermore, in section 1.10.1.6:

All operations on a given mutex occur in a single total order. Each mutex acquisition “reads the value written” by the last mutex release.

And in 30.4.3.1

A mutex object facilitates protection against data races and allows safe synchronization of data between execution agents

This means, acquiring (locking) a mutex sets a one-way barrier that prevents operations that are sequenced after the acquire (inside the protected area) from moving up across the mutex lock.

Releasing (unlocking) a mutex sets a one-way barrier that prevents operations that are sequenced before the release (inside the protected area) from moving down across the mutex unlock.

In addition, memory operations that are released by a mutex are synchronized (visible) with another thread that acquires the same mutex.

In your example, foo_has_been_set is checked in CheckFoo .. If it reads true you know that the value 1 has been assigned to foo by SetFoo , but it is not synchronized yet. The mutex lock that follows will acquire foo , synchronization is complete and the assert cannot fire.

Answer 3

The standard does not directly guarantee that, but you can read it between the lines of [thread.mutex.requirements.mutex].:

For purposes of determining the existence of a data race, these behave as atomic operations ([intro.multithread]).
The lock and unlock operations on a single mutex shall appear to occur in a single total order.

Now the second sentence looks like a hard guarantee, but it really isn't. Single total order is very nice, but it only means that there is a well-defined single total order of of acquiring and releasing one particular mutex . Alone by itself, that doesn't mean that the effects of any atomic operations, or related non-atomic operations should or must be globally visible at some particular point related to the mutex. Or, whatever. The only thing that is guaranteed is about the order of code execution (specifically, the execution of a single pair of functions, lock and unlock ), nothing is being said about what may or may not happen with data, or otherwise.
One can, however, read between the lines that this is nevertheless the very intention from the "behave as atomic operations" part.

From other places, it is also pretty clear that this is the exact idea and that an implementation is expected to work that way, without explicitly saying that it must . For example, [intro.races] reads:

[ Note: For example, a call that acquires a mutex will perform an acquire operation on the locations comprising the mutex. Correspondingly, a call that releases the same mutex will perform a release operation on those same locations.

Note the unlucky little, harmless word "Note:" . Notes are not normative. So, while it's clear that this is how it's intended to be understood (mutex lock = acquire; unlock = release), this is not actually a guarantee.

I think the best, although non-straightforward guarantee comes from this sentence in [thread.mutex.requirements.general]:

A mutex object facilitates protection against data races and allows safe synchronization of data between execution agents.

So that's what a mutex does (without saying how exactly). It protects against data races. Fullstop.

Thus, no matter what subtleties one comes up with and no matter what else is written or isn't explicitly said, using a mutex protects against data races (... of any kind, since no specific type is given). That's what is written. So, in conclusion, as long as you use a mutex, you are good to go even with relaxed ordering or no atomic ops at all. Loads and stores (of any kind) cannot be moved around because then you couldn't be sure no data races occur. Which, however, is exactly what a mutex protects against.
Thus, without saying so, this says that a mutex must be a full barrier.

Answer 4

The answer seem to lie in http://eel.is/c++draft/intro.multithread#intro.races-3

The two pertinent parts are

[...] In addition, there are relaxed atomic operations, which are not synchronization operations [...]

and

[...] performing a release operation on A forces prior side effects on other memory locations to become visible to other threads that later perform a consume or an acquire operation on A. [...]

While relaxed orders atomics are not considered synchronization operations, that's all the standard has to say about them in this context. Since they are still memory locations, the general rule of them being governed by other synchronization operations still applies.

So in conclusion, the standard does not seem to have anything specifically in there to prevent the reordering you described, but the wording as it stands would prevent it naturally.

Edit: Woops, I linked to the draft. The C++11 paragraph covering this is 1.10-5, using the same language.

Answer 5

CheckFoo() cannot cause the program to crash (ie trigger the assert() ) but there is also no guarantee the assert() will ever be executed.

If the condition at the start of CheckFoo() triggers (see below) the visible value of foo will be 1 because of the memory barriers and synchronization between mu.unlock() in SetFoo() and mu.lock() in CheckFoo() .

I believe that is covered by the description of mutex cited in other answers.

However there is no guarantee that the if condition ( foo_has_been_set.load(std::memory_order_relaxed)) ) will ever be true. Relaxed memory order makes no guarantees and only the atomicity of the operation is assured. Consequently in the absence of some other barrier there's no guarantee when the relaxed store in SetFoo() will be visible in CheckFoo() but if it is visible it will only be because the store was executed and then following the mu.lock() must be ordered after mu.unlock() and the writes before it visible.

Please note this argument relies on the fact that foo_has_been_set is only ever set from false to true . If there were another function called UnsetFoo() that set it back to false:

void UnsetFoo() {
  mu.lock();
  foo = 0;
  foo_has_been_set.store(false, std::memory_order_relaxed);
  mu.unlock();
}

That was called from the other (or yet a third) thread then there's no guarantee that checking foo_has_been_set without synchronization will guarantee that foo is set.

To be clear (and assuming foo_has_been_set is never unset):

void CheckFoo() {
  if (foo_has_been_set.load(std::memory_order_relaxed)) {
    assert(foo == 1); //<- All bets are off.  data-race UB
    mu.lock();
    assert(foo == 1); //Guaranteed to succeed.
    mu.unlock();
  }
}

In practice on any real platform on any long running application it is probably inevitable that the relax store will eventually become visible to the other thread. But there is no formal guarantee regarding if or when that will happen unless other barriers exist to assure it.

Formal References:

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3690.pdf

Refer to the notes at the end of p.13 and start of p.14 particularly notes 17 - 20. They are essentially assuring coherence of 'relaxed' operations. Their visibility is relaxed but the visibility that occurs will be coherent and use of the phrase 'happens before' is within the overall principle of program ordering and particularly acquire and release barriers of mutexes. Note 19 is particularly relevant:

The four preceding coherence requirements effectively disallow compiler reordering of atomic operations to a single object, even if both operations are relaxed loads. This effectively makes the cache coherence guarantee provided by most hardware available to C++ atomic operations.

Answer 6

Reordering within the critical section is of course possible:

void SetFoo() {
  mu.lock();
  // REORDERED:
  foo_has_been_set.store(true, std::memory_order_relaxed);
  PAUSE(); //imagine scheduler pause here 
  foo = 1;
  mu.unlock();
}

Now, the question is CheckFoo - can the read of foo_has_been_set fall into the lock? Normally a read like that can (things can fall into locks, just not out), but the lock should never be taken if the if is false, so it would be a strange ordering. Does anything say "speculative locks" are not allowed? Or can the CPU speculate that the if is true before reading foo_has_been_set ?

void CheckFoo() {
    // REORDER???
    mu.lock();
    if (foo_has_been_set.load(std::memory_order_relaxed)) {
        assert(foo == 1);
    }
    mu.unlock();
}

That ordering is probably not OK, but only because of "logic order" not memory order. If the mu.lock() was inlined (and became some atomic ops) what stops them from being reordered?

I'm not too worried about your current code, but I worry about any real code that uses something like this. It is too close to wrong.

ie if the OP code was the real code, you would just change foo to atomic, and get rid of the rest. So the real code must be different. More complicated? ...