Create zombie process

Question

I am interested in creating a zombie process. To my understanding, zombie process happens when the parent process exits before the children process. However, I tried to recreate the zombie process using the following code:

#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main ()
{
  pid_t child_pid;

  child_pid = fork ();
  if (child_pid > 0) {
    exit(0);
  }
  else {
    sleep(100);
    exit (0);
  }
  return 0;
}

However, this code exits right after execute which is expected. However, as I do

ps aux | grep a.out

I found a.out is just running as a normal process, rather than a zombie process as I expected.

The OS I am using is ubuntu 14.04 64 bit

Answer 1

Quoting:

To my understanding, zombie process happens when the parent process exits before the children process.

This is wrong. According to man 2 wait (see NOTES) :

A child that terminates, but has not been waited for becomes a "zombie".

So, if you want to create a zombie process, after the fork(2), the child-process should exit(), and the parent-process should sleep() before exiting, giving you time to observe the output of ps(1).

For instance, you can use the code below instead of yours, and use ps(1) while sleep()ing:

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>

int main(void)
{
    pid_t pid;
    int status;

    if ((pid = fork()) < 0) {
        perror("fork");
        exit(1);
    }

    /* Child */
    if (pid == 0)
        exit(0);

    /* Parent
     * Gives you time to observe the zombie using ps(1) ... */
    sleep(100);

    /* ... and after that, parent wait(2)s its child's
     * exit status, and prints a relevant message. */
    pid = wait(&status);
    if (WIFEXITED(status))
        fprintf(stderr, "\n\t[%d]\tProcess %d exited with status %d.\n",
                (int) getpid(), pid, WEXITSTATUS(status));

    return 0;
}

Answer 2

A zombie or a "defunct process" in Linux is a process that has been completed, but its entry still remains in the process table due to lack of correspondence between the parent and child processes. Usually, a parent process keeps a check on the status of its child processes through the wait() function. When the child process has finished, the wait function signals the parent to completely exit the process from the memory. However, if the parent fails to call the wait function for any of its children, the child process remains alive in the system as a dead or zombie process. These zombie processes might accumulate, in large numbers, on your system and affect its performance.

Below is a c program to creating a Zombie-Process on our system Save this file as zombie.c:

#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main ()
{
  pid_t child_pid;

  child_pid = fork ();
  if (child_pid > 0) {
    sleep (60);
  }
  else {
    exit (0);
  }
  return 0;
}

The zombie process created through this code will run for 60 seconds. You can increase the time duration by specifying a time(in seconds) in the sleep() function.

Compile this program

gcc zombie.c

Now run the zombie program:

./a.out

The ps command will now also show this defunct process, open a new terminal and use the below command to check the defunct process:

aamir@aamir:~/process$ ps -ef | grep a.out
aamir    10171  3052  0 17:12 pts/0    00:00:00 ./a.out
aamir    10172 10171  0 17:12 pts/0    00:00:00 [a.out] <defunct> #Zombie process
aamir    10177  3096  0 17:12 pts/2    00:00:00 grep --color=auto a.out