double fork to avoid zombie process

It is a common mistake to fork a child process without calling waitpid() to wait for the termination of the child process. Without a wait() call, the child process will become a zombie process after its termination because its parent process does not cleanup its process information in the system. A zombie process occupies a pid in the system, decrease the available pids in the system. Zombies are mark as "defunct" if you check the process by the "ps" command.

However, sometimes we do not want the parent process to wait for its child process for a long time. There is a way to achieve both "not create zombie process" and "not wait for the child process to its termination", and the way is to do a double fork.

The idea is simple, when a parent process (say A) want to fork a child process to do "something". Process A does not fork a process to do "something" directly. Process A first forks a child process (say B), and process then forks its child process (say C) to do "something" and process B terminates as soon as process C is created. In this way, process A only has to wait for process B for a short time. In the same time, since it has no parent process (process B is dead), the system will "rechild" process C to the init process. The init process calls wait() for its child process, solving the zombie process problem.

The program looks like

void func()
{
     pit_t pid1;
     pit_t pid2;
     int status;

     if (pid1 = fork()) {
             /* parent process A */
             waitpid(pid1, &status, NULL);
     } else if (!pit1) {
             /* child process B */
             if (pid2 = fork()) {
                     exit(0);
             } else if (!pid2) {
                     /* child process C */
                     execvp("something");
             } else {
                     /* error */
             }
     } else {
             /* error */
     }
}