Re: Failure to optimize?

[Liu Hao via Gcc-help <gcc-help@xxxxxxxxxxx>]

在 2021-01-12 22:20, Jonathan Wakely via Gcc-help 写道:
> I'm not sure about the rules for C, but in C++ the compiler can assume
> there is no race, because the increment is not atomic. If there were
> another access to the variable then a non-atomic store would be a race
> even in the bar1 version.

What about this code:

    // -- beginning of copy-n-pasted code

    char const* foo();

    int cursor = 0;

    char const* bar1() {
        char const* result = foo();
        if (result)
            ++cursor;
        return result;
    }

    char const* bar2() {
        char const* result = foo();
        cursor += !!result;
        return result;
    }

    // -- end of copy-n-pasted code


    #include <atomic>
    #include <thread>
    #include <cstdio>

    char const* foo() {
        static ::std::atomic<char const*> str("meow");
        return str.exchange(nullptr);
    }

    int main() {
      ::std::thread thrs[10];
      for(auto& r : thrs)
        r = ::std::thread(bar1);

      for(auto& r : thrs)
        r.join();

      ::std::printf("cursor = %d\n", cursor);
    }


`foo()` will return non-null for exactly one thread. Increment of `cursor` by that thread is 
sequenced before its termination, which synchronizes with exactly one `join()`, which is sequenced 
before the final read of `cursor`. There is no race in this program, but there would be if `bar2` 
was called in place of `bar1`, where all threads could modify `cursor` concurrently.



--
Best regards,
LH_Mouse

Attachment: OpenPGP_signature
Description: OpenPGP digital signature