I have the following program, just to test if I’m understanding forking and shared memory. In summary it makes shared memory, forks, forks, forks, and in each fork, writes data and exits.

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

#include <sys/shm.h>
#include <sys/ipc.h>

int main() {
    
    key_t key = ftok(".", 'x');
    int shmid = shmget(key, sizeof(int)*4, IPC_CREAT | 0666);
    if (shmid < 0) {
        perror("shmgetn");
        return 1;
    }
    int *arr = shmat(shmid, NULL, 0);
    pid_t c1 = fork();
    if (c1==0) {
        pid_t c2 = fork();
        if (c2==0) {
            pid_t c3 = fork();
            if (c3==0) {
                arr[0] = 10;
            } else {
                arr[1] = 11;
            }
            exit(0);
        } else {
            arr[2] = 12;
        }
        exit(0);
    } else {
        arr[3] = 13;
        wait(NULL);
        wait(NULL);
        wait(NULL);

        for (int i=0; i<4; i++) printf("%d ", arr[i]);
        printf("n");
    }
    exit(0);
}

This prints 0 11 12 13 showing that for some reason the value 10 is never assigned to arr[0].

I expected that every process would reach some call to exit(0). Especially the one after the third fork, I would think should reach the same call to exit(0) that is reached by the second fork.

However, when I add an explicit call to exit inside the third fork’s block (i.e. after c3==0), it then executes as expected.

As I write this, it suddenly occurs to me a guess about why this is happening: Is it because the second child reaches exit before the third, and because it exits then the link from the fourth to the parent is broken?