I am calling fork() twice to create two child processes. I want child process A to do an exec() call and child process B to also do an exec() call. The problem I am having with the given code is that after the first exec() from child process A, the next fork() does not seem to occur and the program exits. I think that it has to do with how exec() overlays the parent process. What I want to accomplish is to call exec() from each of the child processes created by fork().
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/msg.h>
#include <sys/wait.h>
#include <stdlib.h>
#include <unistd.h>
#include <iostream>
int main() {
pid_t cpid_a, cpid_b;
cpid_a = fork();
if(cpid_a < 0) {
std::cout << "Fork failed." << '\n';
return 1;
}
else if(cpid_a == 0) { // code for child process A
execlp("/bin/ls", "ls", NULL);
cpid_b = fork();
if(cpid_b < 0) {
std::cout << "Fork failed." << '\n';
return 1;
}
else if(cpid_b == 0) { // code for child process B
execlp("/bin/ls", "ls", NULL);
}
}
else { // code for parent process
while(wait(NULL) != -1);
}
return 0;
}
execlpreturns, it is only because it failed. You should never have any code after a call to execl that does anything except handle an error. - William Pursellforka second time. - William Pursell