How to run child processes simultaneously? in c - c

Question

So here is my code that will take an int as an command line argument then fork N child processes (That run simultaneously). And then when each child ends, the parent will echo the child that child exit status.
But right now I can only do child by child but not simultaneously. How can I do it?
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <time.h>
int main ( int argc, char *argv[] )
{
int i, pid, ran;
for(i = 0; i < atoi(argv[1]); i++) {
pid = fork();
srand(time(NULL));
ran = (rand() % 10) + 1 ;
if (pid < 0) {
printf("Error");
exit(1);
} else if (pid == 0) {
printf("Child (%d): %d\n", i + 1, getpid());
printf("Sleep for = %d\n", ran);
sleep(ran);
exit(ran);
} else {
int status = 0;
pid_t childpid = wait(&status);
printf("Parent knows child %d is finished. \n", (int)childpid);
}
}
}

Answer 1

You're calling wait() inside of the loop where you're spawning the children, so it won't continue the loop to start the next child until the current one is done.
You need to call wait() outside of the loop in a separate loop:
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <time.h>
int main ( int argc, char *argv[] )
{
int i, pid, ran;
for(i = 0; i < atoi(argv[1]); i++) {
pid = fork();
srand(time(NULL));
ran = (rand() % 10) + 1 ;
if (pid < 0) {
printf("Error");
exit(1);
} else if (pid == 0) {
printf("Child (%d): %d\n", i + 1, getpid());
printf("Sleep for = %d\n", ran);
sleep(ran);
exit(ran);
}
}
for(i = 0; i < atoi(argv[1]); i++) {
int status = 0;
pid_t childpid = wait(&status);
printf("Parent knows child %d is finished. \n", (int)childpid);
}
}