memory_order_relaxed 是否尊重同一线程内的数据依赖性？

Question

Given:

std::atomic<uint64_t> x;

uint64_t f()
{
    x.store(20, std::memory_order::memory_order_relaxed);
    x.store(10, std::memory_order::memory_order_relaxed);
    return x.load(std::memory_order::memory_order_relaxed);
}

Is it ever possible for f to return a value other than 10 , assuming there is only one thread writing to x ? This would obviously not be true for a non-atomic variable, but I don't know if relaxed is so relaxed that it will ignore data dependencies in the same thread?

Answer 1

The result of the load is always 10 (assuming there is only one thread). Even a relaxed atomic variable is "stronger" than a non-atomic variable:

1. as with a non-atomic variable, all threads must agree on a single order in which all modifications to that variable occur,
2. as with a non-atomic variable, that single order is consistent with the "sequenced before" relationship, and
3. the implementation will guarantee that potentially concurrent accesses will somehow sort themselves out into some order that all threads will agree on (and thus satisfy requirement 1). On the other hand, in the case of a non-atomic variable, potentially concurrent accesses result in undefined behaviour.

A relaxed atomic variable can't be used to synchronize different threads with each other, unless accompanied by explicit fences. That's the sense in which it's relaxed, compared with the other memory orderings that are applicable to atomic variables.

For language lawyering, see C++20 [intro.races]/10:

An evaluation A happens before an evaluation B (or, equivalently, B happens after A ) if:

• A is sequenced before B , or [...]

and [intro.races]/15:

If an operation A that modifies an atomic object M happens before an operation B that modifies M , then A shall be earlier than B in the modification order of M . [ Note: This requirement is known as write-write coherence. — end note ]

and [intro.races]/18:

If a side effect X on an atomic object M happens before a value computation B of M , then the evaluation B shall take its value from X or from a side effect Y that follows X in the modification order of M . [ Note: This requirement is known as write-read coherence. — end note ]

Thus, in your program, the store of 20 happens before the store of 10 (since it is sequenced before it) and the store of 10 happens before the load. The write-write coherence requirement guarantees that the store of 10 occurs later in the modification order of x than the store of 20. When the load occurs, it is required to take its value from the store of 10, since the store of 10 happens before it and there is no other modification that can follow the store of 10 in the modification order of x .