使用“sigaction(2)”忽略“SIGCHLD”信号有什么用？

Question

It turns out that we can prevent appearing of a zombie process (ie 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;
}

Answer 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.

Answer 2

The confusion is that you seem to be asuming that " SIGCHLD is ignored by default" means that the default disposition of the signal is SIG_IGN .

That's not the case. The default disposition is SIG_DFL . When you set it to SIG_IGN you're actually changing it. And after you've set it as SIG_IGN , you can revert it back to SIG_DFL , getting the original behavior again.