C++ standard: can relaxed atomic stores be lifted above a mutex lock?

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 s

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.