Question

I’m trying to execute an external program from inside my Linux C++ program.

I’m calling the method system("gedit") to launch an instance of the Gedit editor. However my problem is while the Gedit window is open, my C++ program waits for it to exit.

How can I call an external program without waiting for it to exit?

Answer 1

You will need to use fork and exec

int fork_rv = fork();
if (fork_rv == 0)
{
    // we're in the child
    execl("/path/to/gedit", "gedit", 0);

    // in case execl fails
    _exit(1);
}
else if (fork_rv == -1)
{
    // error could not fork
}

You will also need to reap your child so as not to leave a zombie process.

void reap_child(int sig)
{
    int status;
    waitpid(-1, &status, WNOHANG);
}

int main()
{
    signal(SIGCHLD, reap_child);
    ...
}

In regards to zombie processes, you have a second option. It uses a bit more resources (this flavor forks twice), but the benefit is you can keep your wait closer to your fork which is nicer in terms of maintenance.

int fork_rv = fork();
if (fork_rv == 0)
{
    fork_rv = fork();
    if (fork_rv == 0)
    {
        // we're in the child
        execl("/path/to/gedit", "gedit", 0);

         // if execl fails
        _exit(1);
    }
    else if (fork_rv == -1)
    {
        // fork fails
        _exit(2);
    }

    _exit(0);
}
else if (fork_rv != -1)
{
    // parent wait for the child (which will exit quickly)
    int status;
    waitpid(fork_rv, &status, 0);
}
else if (fork_rv == -1)
{
    // error could not fork
}

What this last flavor does is create a child, which in turns creates a grandchild and the grandchild is what exec’s your gedit program. The child itself exits and the parent process can reap it right away. So an extra fork but you keep all the code in one place.