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