During the period just before Thread A waits on condition variable it must be holding a mutex. The easiest solution is to make sure that Thread B is holding the same mutex at the time it calls notify_all. So something like this:
std::mutex m;
std::condition_variable cv;
int the_condition = 0;
Thread A: {
std::unique_lock<std::mutex> lock(m);
do something
while (the_condition == 0) {
cv.wait(lock);
}
now the_condition != 0 and thread A has the mutex
do something else
} // releases the mutex;
Thread B: {
std::unique_lock<std::mutex> lock(m);
do something that makes the_condition != 0
cv.notify_all();
} // releases the mutex
This guarantees that Thread B only does the notify_all() either before Thread A acquires the mutex or while Thread A is waiting on the condition variable.
The other key here, though, is the while loop waiting for the_condition to become true. Once A has the mutex it should not be possible for any other thread to change the_condition until A has tested the_condition, found it false, and started waiting (thus releasing the mutex).
The point is: what you are really waiting for is for the value of the_condition to become non-zero, the std::condition_variable::notify_all is just telling you that thread B thinks thread A should wake up and retest.
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