Does std::condition_variable::notify_one() or std::condition_variable::notify_all() guarantee that non-atomic memory writes in the current thread prior to the call will be visible in notified threads?
Other threads do:
{
std::unique_lock lock(mutex);
cv.wait(lock, []() { return values[threadIndex] != 0; });
// May a thread here see a zero value and therefore start to wait again?
}
Main thread does:
fillData(values); // All values are zero and all threads wait() before calling this.
cv.notify_all(); // Do need some memory fence or lock before this
// to ensure that new non-zero values will be visible
// in other threads immediately after waking up?
Doesn't notify_all() store some atomic value therefore enforcing memory ordering? I did not clarified it.
UPD: according to Superlokkus' answer and an answer here: we have to acquire a lock to ensure memory writes visibility in other threads (memory propagation), otherwise threads in my case may read zero values.
Also I missed this quote here about condition_variable, which specifically answers my question. Even an atomic variable has to be modified under a lock in a case when the modification must become visible immediately.
Even if the shared variable is atomic, it must be modified under the mutex in order to correctly publish the modification to the waiting thread.