Capturing exit code of n child processes - c

Question

I am trying to fork() 10 child processes in one loop and then in another loop wait() for them to terminate and print their PID along with their exit status code. It cannot be done any other way or using any other function. Two loops/waves and the function wait();
This is what I have tried:
#include <stdio.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <stdlib.h>
#include <unistd.h>
int main()
{
pid_t wait_p, p[10], p_child;
int status;
for (int i = 0; i < 10; i++)
{
p[i] = fork();
}
for (int i = 0; i < 10; i++)
{
switch (p[i])
{
case 0:
p_child = getpid();
exit(p_child % 10);
break;
case -1:
puts("ERROR");
break;
default:
wait_p = wait(&status);
printf("Child with PID: %d", wait_p);
if (WIFEXITED(status))
printf(" terminated with STATUS: %d\n", WEXITSTATUS(status));
break;
}
}
return (EXIT_FAILURE);
}
This code will execute an endless count of child processes. It must print only the first original(issued by THE one parent) 10. What am I doing wrong?

Answer 1

You have to handle the child processes directly in your first loop:
for (int i = 0; i < 10; i++)
{
p[i] = fork();
if (p[i] == 0) {
p_child = getpid();
exit(p_child % 10);
} else if (p[i] == -1) {
perror("fork");
}
}
and then wait for them in the second loop
for (int i = 0; i < 10; i++)
{
wait_p = wait(&status);
printf("Child with PID: %d", wait_p);
if (WIFEXITED(status))
printf(" terminated with STATUS: %d\n", WEXITSTATUS(status));
}
You cannot handle the case, that fork() returns in the child process (yielding 0 as return value), in your second loop, otherwise each child process in the first loop keeps forking more child processes.