std :: condition_variable :: notify_one（）多次调用而无需上下文切换

Question

How many waiting threads will wake up in this example:

1st thread :

void wakeUp2Threads()
{
    std::unique_lock<std::mutex> lock(condvar_mutex);

    condvar.notify_one();
    condvar.notify_one();
}

2nd thread :

{
    std::unique_lock<std::mutex> lock(condvar_mutex);

    condvar.wait(lock); <- 2nd thread has entered here before 1st thread entered wakeUp2Threads.
}

3rd thread (the same as 2nd):

{
    std::unique_lock<std::mutex> lock(condvar_mutex);

    condvar.wait(lock); <- 3rd thread has entered here before 1st thread entered wakeUp2Threads.
}

Is there any guarantee that in this example both notifications will be delivered to different threads, not the same thread several times?

Ie what does notify_one() mean:

1) notify one thread, no matter has it been already notified (but has not been woken up yet), or not. (* see note)
or
2) notify one thread, but only this one, which has not been notified yet.

(*) Pay attention! I'm not talking here about scenario "waiting thread had already been notified somewhere in the past, woken up, done some stuff and entered condvar.wait() again" - of course, in this case several notify_one() routines can wake up the same thread over and over again.

I'm talking about another case :

notify_one() has notified waiting thread about wake-up, but BEFORE this waiting thread has received time slot form the kernel scheduler and continued execution - another notify_one() has been called again. Is it possible this second notification will be delivered to the same thread again, while it hasn't been woken up from first notification yet?

Answer 1

The notify_one call atomically unblocks one thread. That means that when called a second time it can't unblock the same thread as it's no longer blocked.

This is specified in the standard at section 30.5/3 and 30.5.1/7.