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
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;
}