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Comment by rigtorp | Hacker News Reader

There's even more discussion on the lock memory ordering on Stackoverflow: https://stackoverflow.com/questions/61299704/how-c-standard-...

Taking a lock only needs to be an acquire operation and a compiler barrier for other lock operations. Using seq_cst or acq_rel semantics is stronger than needed. From my reading and discussions with people from WG21 the current argument for why taking a lock only requires acq semantics is that a compiler optimization that transforms a non-deadlocking program into a potentially deadlocking program is not allowed. There's an interesting twitter thread where we discuss this I can't find anymore :(.

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That is an amazing thread. The fact that C++ apparently allows optimizing

    #include <stdio.h>
    
    int stop = 1;
    
    void maybeStop() {
        if(stop)
            for(;;);
    }
    
    int main() {
        printf("hello, ");
        maybeStop();
        printf("world\n");
    }

into

    int main() {
        printf("hello, world\n");
    }

(as Clang does today) does not inspire confidence about disallowing moving the loop in the other example. If the compiler is allowed to assume that this loop terminates, why not the lock loop?

Maybe there is a reason, but none of this inspires confidence.

I don't remember the exact wording, but the standard explicitly makes an exception for the always terminating assumptions, for loops accessing atomic variables or having side effects (i.e. volatile or I/O).

The standard says that a thread must eventually terminate, do an atomic operation or do IO. So the while(lock.exchange(true)); loop is different.

Also keep in mind that C++11 specifies std::mutex::lock() to have acquire semantics and unlock() to have release semantics on the lock object. In order for std::mutex to actually work the reordering of m1.unlock(); m2.lock(); to m2.lock(); m1.unlock(); must be disallowed. But since m1 and m2 are separate objects m1.unlock() has no happens before relationship with m2.lock(). This seems to be a problem in the C++11 memory model. The arguments I have heard from some WG21 people is that there is no problem since transforming a wellformed terminating program into a non-terminating program is not allowed. I can't find the wording in the C++ standard that asserts this. But oh well, it works right now on gcc/llvm/msvc.

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