调用 fork 两次避免僵尸进程

Avoid zombie processes by calling fork twice

/*
* Avoid zombie processes by calling fork twice.
* APUE-2e 程序清单8-5
*/
#include <unistd.h>
#include <sys/wait.h>
#include <stdio.h>
#include <stdlib.h>

int main()
{
pid_t pid;
if( (pid = fork()) < 0 )
{
printf("fork error.\n");
exit(-1);
}
else if(pid == 0) /* first child */
{
if( (pid = fork()) < 0 )
{
printf("fork error.\n");
exit(-1);
}
else if(pid > 0)
{
exit(0);
}

/* We're the second child; our parent becomes init as soon as our real parent exits. */
printf("second child, parent pid = %d\n", getppid());
/* ---------------handle tasks--------------- */
exit(0);
}

if(waitpid(pid, NULL, 0) != pid) /* wait for first child */
{
printf("waitpid error.\n");
exit(1);
}
printf("parent, first child pid = %d\n", pid);
/* ---------------handle tasks--------------- */

exit(0);
}

注意

这个代码并不是通用场景的避免僵尸进程的办法，它的场景是这样的：

一个进程要创建一个进程，两个进程同时处理任务，谁也不耽误谁。如果直接用子进程充当第二个进程的角色，那么问题是这样的：如果父进程处理时间长，子进程处理时间短，那么如果父进程不 wait() 处理的话，子进程就会成为僵尸进程，但如果父进程 wait() 子进程的话，父进程就会阻塞，所有有个方法就是让自己尽快推出，任务让子进程的子进程来处理。

这个场景的 APUE 的原文描述：If we want to write a process so that it forks a child but we don't want to wait for the child to complete and we don't want the child to become a zombie until we terminate, the trick is to call fork twice.