How can I prevent zombie child processes?

Question

I am writing a server that uses fork() to spawn handlers for client connections. The server does not need to know about what happens to the forked processes – they work on their own, and when they're done, they should just die instead of becoming zombies. What is an easy way to accomplish this?

Neo

This, probably, is the easiest way: #7172222 – Razorbill

@Razorbill Oh, nice. Do you think the questions are similar enough to flag this as a duplicate? – Neo

Answer 1

There are several ways, but using sigaction with SA_NOCLDWAIT in the parent process is probably the easiest one:

struct sigaction sigchld_action = {
  .sa_handler = SIG_DFL,
  .sa_flags = SA_NOCLDWAIT
};
sigaction(SIGCHLD, &sigchld_action, NULL);

Neo

Even easier: signal(SIGCHLD, SIG_IGN); – Benedict

Answer 2

Use double forks. Have your children immediately fork another copy and have the original child process exit.

http://thinkiii.blogspot.com/2009/12/double-fork-to-avoid-zombie-process.html

This is simpler than using signals, in my opinion, and more understandable.

void safe_fork()
{
  pid_t pid;
  if (!pid=fork()) {
    if (!fork()) {
      /* this is the child that keeps going */
      do_something(); /* or exec */
    } else {
      /* the first child process exits */
      exit(0);
    }
  } else {
    /* this is the original process */  
    /* wait for the first child to exit which it will immediately */
    waitpid(pid);
  }
}

Fridlund

However, if someone wants to see a process tree or something like that, this will mess it up, right? – Neo

Not sure, but it seems likely. – Fridlund

On POSIX systems, the child's parent will be changed to INIT (PID = 1), so yeah, the process tree will not be what you might have expected – Iterative

Is this correct? The fact that the second child process is not becoming a zombie is reliant on the fact that the first child exits immediately. What if do_something() does not do anything? – Evslin

sure you can add some sort of gating mechanism if needed. – Fridlund

And it works on a per process level and not global. Global configurations are as evil as global variables. – Leveloff

Is that correct that processes don't become zombies but each time the safe_fork() is executed, there's another process in ps which only ends when the parent of all, terminates? – Fenny

@AbelRomeroPérez They are re-parented to init which cleans them up as soon as they're complete. The short-lived process is cleaned up by the parent process immediately. – Fridlund

@Fridlund and how can be there's a new process in ps everytime my own implementation of safe_fork() is executed? :( – Fenny

Answer 3

How to get rid of zombie processes?

you can’t kill the zombie process with SIGKILL signal as you kill a normall process, As the zombie process can’t recive any signal. so having a good habit is very important.

Then when programming, how to get rid amount of zombie processes? According to the above description, the child process will send SIGCHLD signals to the parent process when its dies. by default, this signal is ignored by system, so the best way is that we can call wait() in the signal processing function, which could avoid the zombie stick around in the system. see more about this: http://itsprite.com/how-to-deep-understand-the-zombie-process-in-linux/

Dextrin