Waitpid blocks forever

Question

I have a little confusion with waitpid function:

int main(int argc, char** argv) {
    if (pid_t pid = fork()) {
        setpgid(pid, pid);
        waitpid(pid, NULL, 0);
    } else {
        setpgid(0, 0);
        char *args[] = {"man", "2", "pipe", NULL};
        execvp(args[0], args);
    }

    return 0;
}

In this example program just blocks and to nothing. The same story goes with less and every program which waits for you to press 'q'. On the other hand if I replace man with say "ls" - everything works perfectly: it prints content of a directory.

char *args[] = {"ls", NULL};

So what is it so special about man?

EDIT: I forget to say that I need the process to be the leader of a group. In case I remove setpgid() everything is also OK:

int main(int argc, char** argv) {
    if (pid_t pid = fork()) {
        waitpid(pid, NULL, 0);
    } else {
        char *args[] = {"man", "2", "pipe", NULL};
        execvp(args[0], args);
    }

    return 0;
}

Are you creating your own shell? I don't know exactly what processes groups are for, but it in the `setpgid()` man page it seems they are used by a shell to regroup processes of a same job, so maybe it's bad idea to try to override this unless you are writing a shell. — jdarthenay, Apr 07 '16 at 19:57
Yeah I'm writing shell with pipes. I need process group to send signals to all the programs in the pipe. — Alexandr, Apr 07 '16 at 20:03
Can you compile that with a C compiler? (The declaration in the `if` condition) — Daniel Jour, Apr 07 '16 at 20:13
@ChrisDodd `error: unexpected type name 'pid_t': expected expression` says `cc -std=c11 file.c` where `cc` is `clang 3.7.1` — Daniel Jour, Apr 07 '16 at 20:28
Hmm, I guess I'm looking at a newer draft of the spec then. Not sure exactly when it was added. — Chris Dodd, Apr 07 '16 at 20:31

score 2 · Accepted Answer · answered Apr 07 '16 at 20:04

You are putting the child process in its own process group. The session's controlling terminal has only one foreground process group at any given time, and you do nothing to put the child's process group in the foreground. Therefore, it does not receive keyboard input, and its standard output is not displayed on the terminal.

Because the command you are exec'ing in the child waits for keyboard input that will never come while that command is in the background, that command never exits, and waitpid() never collects it.

Unless you have a specific reason for putting the child in a different process group, you probably should not do so.

If you do have need of putting the child into its own process group, you should be using `tcsetpgrp` to control the terminal's foreground process group. — Chris Dodd, Apr 07 '16 at 20:28

Waitpid blocks forever

1 Answers1