Calling pthread_cond_signal without locking mutex

Question

I read somewhere that we should lock the mutex before calling pthread_cond_signal and unlock the mutex after calling it:

The pthread_cond_signal() routine is used to signal (or wake up) another thread which is waiting on the condition variable. It should be called after mutex is locked, and must unlock mutex in order for pthread_cond_wait() routine to complete.

My question is: isn't it OK to call pthread_cond_signal or pthread_cond_broadcast methods without locking the mutex?

caf caf · Accepted Answer · 2010-12-31T03:38:23

If you do not lock the mutex in the codepath that changes the condition and signals, you can lose wakeups. Consider this pair of processes:

Process A:

pthread_mutex_lock(&mutex);
while (condition == FALSE)
    pthread_cond_wait(&cond, &mutex);
pthread_mutex_unlock(&mutex);

Process B (incorrect):

condition = TRUE;
pthread_cond_signal(&cond);

Then consider this possible interleaving of instructions, where condition starts out as FALSE:

Process A                             Process B

pthread_mutex_lock(&mutex);
while (condition == FALSE)

                                      condition = TRUE;
                                      pthread_cond_signal(&cond);

pthread_cond_wait(&cond, &mutex);

The condition is now TRUE, but Process A is stuck waiting on the condition variable - it missed the wakeup signal. If we alter Process B to lock the mutex:

Process B (correct):

pthread_mutex_lock(&mutex);
condition = TRUE;
pthread_cond_signal(&cond);
pthread_mutex_unlock(&mutex);

...then the above cannot occur; the wakeup will never be missed.

(Note that you can actually move the pthread_cond_signal() itself after the pthread_mutex_unlock(), but this can result in less optimal scheduling of threads, and you've necessarily locked the mutex already in this code path due to changing the condition itself).

Calling pthread_cond_signal without locking mutex

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