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?

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);

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);
  }
}

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/