Processes and semaphores

Question

So i was experimenting on how to use fork and semaphores for a homework and it seems everytime i run the program fork always returns a number >0, while what i wanted was to first have several processes be made then be stopped using semaphores and then have some of them restart again.

#include <unistd.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/sem.h>
#include <semaphore.h>

int     main(int argc,char *argv[])
{
int i,count;
 count = 0;
 pid_t  *Pc;
 Pc=(pid_t *) malloc((argc-2)*sizeof(pid_t));
 sem_t *sem;
  sem = (sem_t *) malloc((argc-2)*sizeof(sem_t));
     for (i = 0; i <= argc-2; i++){
      sem_init(&sem[i], 0, 0);
 }
     for (i =0; i<=argc-2; i++){
                    Pc[i] = fork();
                    if (Pc[i] == 0){
                   printf(" child");
                    sem_wait(&sem[i]);
                    printf("Experiment was a success!");
          }
           if (Pc[i]>0){
             printf("Parent");
}
    }
        for (i =0; i<=argc-2; i++){
           if (Pc[i] > 0)
            count++;
        }

    for (i= 0; i<=3; i++){
            if  ( count ==  argc-2){
            sem_post(&sem[i]);
 }
    }

}

nameofprogram 1 2

prints: Parent Child Parent Child

Remember that stdout typically is line buffered. If your messages don't show up as expected, add some "\n", as in printf(" child\n"), or use fflush, as in fflush(stdout). — Thomas Padron-McCarthy, Nov 27 '14 at 16:52
Messages come out but it's always Parent what is being printed. Child or Experiment was a success! never are. Which, makes me wonder what is wrong with it :/ — Geo Ha, Nov 27 '14 at 16:55
So after adding new lines and running it with nameofprogram 1 2 i got Parent Child Parent Child yet again the message after child processes should be restarting didn't show up... i am suspecting Parent is getting printed first cause the parent process is faster, not sure though. — Geo Ha, Nov 27 '14 at 17:11

score 2 · Answer 1 · edited May 23 '17 at 12:27

2

You need to read the man page for sem_init(). The type of semaphore you are creating right now is not shared across processes. This requires a non-trivial change to your program, because you also need to set up shared memory. Refer to this question for a lengthy explanation of how to make your program work.

edited May 23 '17 at 12:27

Community

1
1

answered Nov 27 '14 at 18:10

JS1

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score 2 · Answer 2 · edited May 23 '17 at 12:05

When a program calls fork, a new process is created with a new exact copy of the memory space. This mean that sem in your child process is not the same as sem in your parent process. So, when you call sem_post, you child process can not be notified of the change and get out of the wait function.

To solve this, you have different possibilities:

Create a shared memory which can be read by all your processes and create this semaphore in this shared memory, as already suggested.
Use named semaphores with sem_open. This kind of semaphore can be shared across different processes as it work like a file handle. This seems to be an easier way if you only need a shared semaphore (example here). You will have to generate a unique name for each semaphore in your array (may be only one semaphore on which you call sem_post multiple times would be enough for your use).
Keep your semaphores and use threads instead of processes (but I guess your homework is about processes so this may not be an option for you)

Processes and semaphores

2 Answers2