posix semaphores posting for more than one thread at a time

Hello All,

I would like to know can we post more than one thread at a time.

I have code like below

#include <stdio.h>
#include <unistd.h>
#include <errno.h>
#include <stdlib.h>
#include <string.h>

#include <pthread.h>
#include <semaphore.h>

#define MAX_MSG_LEN 256

sem_t sem1;

char msg1[MAX_MSG_LEN] = "1";

void *thrdFun1(void *arg);

int main()
{
     pthread_t thrd1,thrd2,thrd3,thrd4;
     char argmsg1[] = "Thread1: ";
     char argmsg2[] = "Thread2: ";
     char argmsg3[] = "Thread3: ";
     char argmsg4[] = "Thread4: ";
     int res;
     int thNum;

     res = sem_init(&sem1,0,0);
     res = pthread_create(&thrd1, NULL, thrdFun1, argmsg1);
     res = pthread_create(&thrd2, NULL, thrdFun1, argmsg2);
     res = pthread_create(&thrd3, NULL, thrdFun1, argmsg3);
     res = pthread_create(&thrd4, NULL, thrdFun1, argmsg4);

     while(1)
     {
          printf("Enter message to end to thread \n");
          fgets(msg1,MAX_MSG_LEN,stdin);
          sem_post(&sem1);
     }

     return 0;          /* writing the comments */
}

void *thrdFun1(void *arg)
{
     while(1)
     {
          sem_wait(&sem1);
          printf("Iam %s message is %s\n",arg,msg1);
     }
}

here it does accept and sem_post serially to the thread than invoking all the threads at a time.

Do this kind of behaviour can be achived ( posting to all threads)?

your comments would help in understanding .

regards

Hi,

I think that a semaphore is the wrong synchronization primitive to use for this situation. I think you actually want a condition variable. They mean "form an orderly queue behind a single mutex, until the condition is met, then the service will either let one in at a time, or it's a free-for-all" and I think you want the free-for-all option.

see man 3 pthread_cond_init

Then to wake up all the threads waiting on the condition variable, you would use pthread_cond_broadcast like so:

#include <stdio.h>
#include <unistd.h>
#include <errno.h>
#include <stdlib.h>
#include <string.h>

#include <pthread.h>
#include <semaphore.h>

#define MAX_MSG_LEN 256

pthread_mutex_t mut = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;

char msg1[MAX_MSG_LEN] = "1";

void *thrdFun1(void *arg);

int main()
{
     pthread_t thrd1,thrd2,thrd3,thrd4;
     char argmsg1[] = "Thread1: ";
     char argmsg2[] = "Thread2: ";
     char argmsg3[] = "Thread3: ";
     char argmsg4[] = "Thread4: ";
     int res;
     int thNum;

     res = pthread_create(&thrd1, NULL, thrdFun1, argmsg1);
     res = pthread_create(&thrd2, NULL, thrdFun1, argmsg2);
     res = pthread_create(&thrd3, NULL, thrdFun1, argmsg3);
     res = pthread_create(&thrd4, NULL, thrdFun1, argmsg4);

 
     while(1)
     {
          pthread_mutex_lock(&mut);
          printf("Enter message to end to thread \n");
          fgets(msg1,MAX_MSG_LEN,stdin);
          pthread_cond_broadcast(&cond);
          pthread_mutex_unlock(&mut);
     }

     return 0;          /* writing the comments */
}

void *thrdFun1(void *arg)
{
    pthread_mutex_lock(&mut);
    /* The condition you're waiting on goes here.  It's almost certainly not a while(1) loop*/
    while (1) {
        pthread_cond_wait(&cond, &mut);
    }
    pthread_mutex_unlock(&mut);
    printf("Iam %s message is %s\n",arg,msg1);
}

I haven't compiled and tried this, so there might be some minor errors (though post back here and I'll see if I can figure them out if you're stuck) and this is also likely to produce garbled output to the terminal, but that's proof that it works as that means more than one thread woke up and started writing to the terminal. Of course, you might get nice output too, there is such a small amount of work for each thread (especially if you're on a single core system), it's possible that each thread will complete in a single scheduler timeslice. That's what sucks about threading - I can't tell you easily ahead of time what kind of output you're going to get, without writing another 100 or so lines of code!


Hth,
Michael

Of course, I've just realised that this program isn't the best test case... I can't guarantee that all the threads have initialised and are waiting on the condition to be met (that's probably the condition you want to wait on in the threads' worker function). And the while(1) loop in the main thread is going to get stuck forever, so you probably want to get rid of that.

I don't really have time to fix it right now, but as it's an example, I think it at least points you in the right direction. But like I said in the previous post; post any follow-up questions here and I'll see if I can get them resolved.

Michael