Semaphore cleanup in Linux c

Question

When I create a shared memory (c program in Linux) I delete it with shmctl(shmid, IPC_RMID, 0) and everything looks fine when I'm using ipcs -m to check if there are any remaining shared memory segments. But I'm wondering how can I delete my semaphores that I've created right before the program is terminated because when I'm using ipcs -s I see both of my semaphores right there, result:

------ Semaphore Arrays --------
key        semid      owner      perms      nsems     
0x6b014021 0          benjamin   600        1         
0x6c014021 1          benjamin   600        1    

Thanks.

Answer 1

You can use semget and semctl after setting KEY to the right value returned by ipcs -s:

    #define KEY 0x...

    int id, rc;

    id = semget(KEY, 1, IPC_STAT);
    if (id < 0)
    {
        perror("semget"); 
        exit(1);
    }

    rc = semctl(id, 1, IPC_RMID); 
    if (rc < 0)
    {
        perror("semctl"); 
        exit(1);
    }

Or use directly semctl with id returned by ipcs -s:

rc = semctl(id, 1, IPC_RMID); 
if (rc < 0)
{
    perror("semctl"); 
    exit(1);
}

Full C program:

    #include <stdio.h>
    #include <stdlib.h>
    #include <unistd.h>
    #include <sys/types.h>
    #include <sys/sem.h>
    #include <errno.h>

    int main(int argc, char **argv){

            int id, rc;

            id = atoi(argv[1]);

            printf("id=%d\n", id);
            rc = semctl(id, 1, IPC_RMID); 
            if (rc < 0)
            {
                perror("semctl"); 
                exit(1);
            }

            exit(0);
        }

Execution:

$ ipcs -s

------ Tableaux de sémaphores --------
clef       semid      propriétaire perms      nsems           
0x00001111 393221     pifor      666        1         

$ ./rsem 393221
id=393221
$ ipcs -s

------ Tableaux de sémaphores --------
clef       semid      propriétaire perms      nsems