Semaphore cleanup in Linux c - 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