Why waitpid return -1 when run in debugger?

Question

I'm using fork to create a process on a Mac platform, and wait for the child process to finish in the parent process. But the waitpid return -1 and errno is 4 (EINTR).

The example code, which can reproduce this problem, is as follows:

#include <iostream>
#include <sys/types.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <assert.h>
int main(int argc, const char *argv[])
{
    pid_t pid = ::fork();
    if (pid == 0)
    {
        return 0;
    }

    int s = 0;
    if (::waitpid(pid, &s, 0) == -1)
    {
        printf("The errno is :%d\n", errno); // <<<The errno is 4(EINTR) in my machine.
        assert(false);  // <<<<This will be hit if run in debugger.
    }
    return 0;
}

When I run this code in GDB or LLDB the assert will always be hit. If not run in debugger it will not return -1.

I think there is something I don't understand about how debugger or fork/waitpid works. So can anyone explain why this happen?

The output of your printf would be more telling if you included it in your question (and tailed it with a '\n'to flush it. Pretty sure it is going to be EINTR; an interrupted system call. — WhozCraig, Feb 04 '14 at 13:53
Seams Xcode show the `errno` value incorrectly, the actual `errno` value is 4(`EINTR`). I have correct it in this post. — ZijingWu, Feb 04 '14 at 13:59
`waitpid` is restartable if EINTR is returned. Structure a loop accordingly. I've had to do the same on my Mac. — WhozCraig, Feb 04 '14 at 14:05
A SIGCHLD signal is sent to the parent process when the child process terminates. That interrupts the waitpid() call. — Martin R, Feb 04 '14 at 14:14
The only post on the theme with really good public reaction is: http://stackoverflow.com/a/2708512/715269 — Gangnus, Feb 04 '14 at 20:51

bu11d0zer · Answer 1 · 2018-07-23T21:11:25.653

The debugger is sending a signal to the process. As WhozCraig said, these should be handled by detecting EINTR and calling waitpid again in a loop. Be aware that when running the child process in the debugger, the debugger likely changes the parent process to the debugger's PID, so waitpid on the child PID will fail thinking the child process has exited, even though it has not. When the debugger exits, it restores the PPID of the child to the original parent's PID.

Why waitpid return -1 when run in debugger?

1 Answers1