What is the use of ignoring `SIGCHLD` signal with `sigaction(2)`?

问题

It turns out that we can prevent appearing of a zombie process (i.e. the one whose parent doesn't wait() for it to _exit()) by specifying SIGCHLD signal to be ignored with sigaction() by its parent. However, it seems like SIGCHLD is ignored by default anyway. How come does this work?

int main (void) {
    struct sigaction sa;
    sa.sa_handler = SIG_IGN; //handle signal by ignoring
    sigemptyset(&sa.sa_mask);
    sa.sa_flags = 0;
    if (sigaction(SIGCHLD, &sa, 0) == -1) {
        perror(0);
        exit(1);
    }
    int pid = fork();
    if (pid == 0) { //child process
        _exit(0);
    }
    do_something(); //parent process
    return 0;
}

回答1:

The default behavior of SIGCHLD is to discard the signal, but the child process is kept as a zombie until the parent calls wait() (or a variant) to get its termination status.

But if you explicitly call sigaction() with the disposition SIG_IGN, that causes it not to turn the child into a zombie -- when the child exits it is reaped immediately. See https://stackoverflow.com/a/7171836/1491895

The POSIX way to get this behavior is by calling sigaction with handler = SIG_DFL and flags containing SA_NOCLDWAIT. This is in Linux since 2.6.

来源：https://stackoverflow.com/questions/40601337/what-is-the-use-of-ignoring-sigchld-signal-with-sigaction2