Fork, execlp and kill. Zombie process

Question

I have a c program that executes another process (bash script) with fork and execlp. When I want to kill this process it moves to a zombiee state. Why is this??

Create a process:

    switch (PID_BackUp= fork())
    {
      case -1:
          perror("fork");
          printf("An error has occurred launching backup.sh script!\n");
          break;

      case 0:
          execlp("/usr/local/bin/backup.sh", "/usr/local/bin/backup.sh", NULL, NULL, NULL);
          break;

      default:
          printf("backup.sh script launched correctly! PID: %d\n", PID_BackUp);
          break;
    }

Kill a process:

    if(kill(PID_BackUp,SIGKILL)==-1)
        fprintf(stderr, "No se ha podido matar el programa: %d\n", PID_BackUp);
    else
        printf("\nProceso con identificador %d, se ha abortado\n", PID_BackUp);

So at this point the process moves to a zombie state. What I am doing wrong?

Basile Starynkevitch Basile Starynkevitch · Accepted Answer · 2017-05-26T06:49:04

You should call waitpid(2) or some other waiting system call (wait4(2)...) to avoid having zombie processes. BTW you may also handle the SIGCHLD signal to be notified when a child process has terminated (or changed state).

Generally, you'll better first kill(2) a process with SIGTERM (and only later, with SIGKILL) since that process -your backup.sh script- could handle correctly (using trap builtin in the shell script) the signal (but SIGKILL cannot be caught, and the backup could e.g. leave clutter or useless files on the disk). See also signal(7) & signal-safety(7)

Read Advanced Linux Programming (we can't explain what is taught by that freely available book in several chapters in only a few paragraphs).

BTW, better use perror(3) or strerror(errno) in messages when a system call fails.

Fork, execlp and kill. Zombie process

1 Answers