[SOLVED] Unlocking the condition variable mutex twice?

Issue

I'm looking at the following snippets:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <pthread.h>

pthread_mutex_t mtx = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cvar;
char buf[25];
void*thread_f(void *);

int main(){

    pthread_t thr; 
    pthread_cond_init(&cvar,NULL);
    pthread_mutex_lock(&mtx);
    pthread_create(&thr,NULL,thread_f,NULL);
    pthread_cond_wait(&cvar,&mtx);
    pthread_mutex_unlock(&mtx);
    ...join and destroy thread...
}

and thread_f:

void* thread_f(void* argp){

    pthread_mutex_lock(&mtx);
    strcpy(buf,"test");
    pthread_cond_signal(&cvar);
    pthread_mutex_unlock(&mtx); //why?
    pthread_exit(NULL);
}

My understanding of the above code is, that main grabs the lock, creates the thread and runs thread_f which blocks. Then, main signals the wait on the condition variable cvar and unlocks the mutex. Then, thread_f unblocks, strcpys and signals cvar to wake up. Then, both main and thread_f unlock the mutex.

Is this the actual behaviour, or am I missing something somewhere? If it's the actual behaviour, isn't unlocking the already unlocked mutex UB, therefore should one of he mutex unlocks in the end be removed?

Thanks.

Solution

What this code is doing is valid.

When the thread calls pthread_cond_signal it doesn't immediately cause the main thread to grab the mutex, but wakes up the main thread so it can attempt to lock it. Then once the thread releases the mutex, the main thread is able to lock it and return from pthread_cond_wait.

Answered By - dbush

Answer Checked By - Dawn Plyler (WPSolving Volunteer)