How to initialise a binary semaphore in C

Question

In the man page it appears that even if you initialise a semaphore to a value of one:

sem_init(&mySem, 0, 1);

It could still be incremented to a value greater than 1 with multiple calls to

sem_post(&mySem);

But in this code example the comment seems to think differently:

sem_init(&mutex, 0, 1);      /* initialize mutex to 1 - binary semaphore */

Is it possible to initialise a strictly binary semaphore in C?

Note: The reason for doing this instead of using a mutex in this case is the sem_post and sem_wait may be called by different threads.

Dietrich Epp · Accepted Answer · 2020-06-02 19:32:36Z

12

If you want a strictly binary semaphore on Linux, I suggest building one out of mutexes and condition variables.

struct binary_semaphore {
    pthread_mutex_t mutex;
    pthread_cond_t cvar;
    bool v;
};

void mysem_post(struct binary_semaphore *p)
{
    pthread_mutex_lock(&p->mutex);
    if (p->v)
        abort(); // error
    p->v = true;
    pthread_cond_signal(&p->cvar);
    pthread_mutex_unlock(&p->mutex);
}

void mysem_wait(struct binary_semaphore *p)
{
    pthread_mutex_lock(&p->mutex);
    while (!p->v)
        pthread_cond_wait(&p->cvar, &p->mutex);
    p->v = false;
    pthread_mutex_unlock(&p->mutex);
}

edited Jun 2, 2020 at 19:32

answered Sep 20, 2011 at 0:24

Dietrich Epp

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11 Comments

Bill DeRose Over a year ago

What exactly is the point of the while loop in mysem_wait? Won't a simple if-statement do the job since pthread_cond_wait will reaquire the mutex after being signaled and pthread_cond_signal only unblocks a single thread waiting on the condition?

Dietrich Epp Over a year ago

@BillDeRose: Condition variables are subject to spurious wakeups and stolen wakeups. You must use them in a loop if you want to check for a condition.

Mquinteiro Over a year ago

@DietrichEpp I can't understand, supose that you call wait before calling post. p->mutex is locked so if you call sem_post the thread will lock in sem_post and everything will locked (I'm sure that I'm wrong but I don't get it)If pthread_cond_signal unlock the mutex, why do you use additional mutex unlock

Dietrich Epp Over a year ago

@Mquinteiro: Read the manual page for pthread_cond_wait. It unlocks the mutex while it is waiting. I don’t see any extra mutex unlock here, each call to lock() is paired with a call to unlock() at the end of the function.

Mquinteiro Over a year ago

@DietrichEpp that means that when is signaled and wait the finish p->mutex is locked again?

|

Ahmed Fawad · Accepted Answer · 2022-06-26 19:49:49Z

0

you can create a wrapper function for sem_post that will check the semaphore value before sem_post().

example:

int semaphore_give(sem_t *sem)
{
    int value, ret;

    if (!sem)
        return -1;
    
    if (sem_getvalue(sem, &value))
        return -1;

    if (value == 0)
        return sem_post(sem);
    return -1;
}

answered Jun 26, 2022 at 19:49

Ahmed Fawad

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Comments

Peter · Accepted Answer · 2022-05-20 09:48:13Z

-1

I don't like the implementation of semaphores in Linux but have to use it. To make it simple I decrement sem value till zero with sem_trywait if it is returned one on the first check:

if (sem_trywait(pProcess->sem_id) < 0 && errno != EINTR ) {
    return ERROR;
}
else {
    while (sem_trywait(pProcess->sem_id) == 0);
    return OK;
}

Be careful if sem_post is called too often. You can stuck in the while() clause. In my case it's called no more than once per second.

answered May 20, 2022 at 9:48

Peter

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