igor igor - 17 days ago 13
Linux Question

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

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;
}

Answer

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 http://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.

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