如何防止僵尸子进程？

I am writing a server that uses fork() to spawn handlers for client connections. The server does not need to know about what happens to the forked processes – they work on their own, and when they're done, they should just die instead of becoming zombies. What is an easy way to accomplish this?

我正在编写一个服务器，它使用fork（）为客户端连接生成处理程序。服务器不需要知道分叉进程会发生什么 - 它们自己工作，当它们完成时，它们应该只是死而不是变成僵尸。什么是实现这一目标的简单方法？

3 个解决方案

#1

There are several ways, but using sigaction with SA_NOCLDWAIT in the parent process is probably the easiest one:

有几种方法，但在父进程中使用SA_NOCLDWAIT的sigaction可能是最简单的方法：

struct sigaction sigchld_action = {
  .sa_handler = SIG_DFL,
  .sa_flags = SA_NOCLDWAIT
};
sigaction(SIGCHLD, &sigchld_action, NULL);

#2

Use double forks. Have your children immediately fork another copy and have the original child process exit.

使用双叉。让您的孩子立即分叉另一个副本并让原始子进程退出。

http://thinkiii.blogspot.com/2009/12/double-fork-to-avoid-zombie-process.html

This is simpler than using signals, in my opinion, and more understandable.

在我看来，这比使用信号更简单，更容易理解。

void safe_fork()
{
  pid_t pid;
  if (!pid=fork()) {
    if (!fork()) {
      /* this is the child that keeps going */
      do_something(); /* or exec */
    } else {
      /* the first child process exits */
      exit(0);
    }
  } else {
    /* this is the original process */  
    /* wait for the first child to exit which it will immediately */
    waitpid(pid);
  }
}

#3

How to get rid of zombie processes?

如何摆脱僵尸进程？

you can’t kill the zombie process with SIGKILL signal as you kill a normall process, As the zombie process can’t recive any signal. so having a good habit is very important.

当你杀死一个正常的进程时，你不能用SIGKILL信号杀死僵尸进程，因为僵尸进程无法回复任何信号。所以养成良好的习惯非常重要。

Then when programming, how to get rid amount of zombie processes? According to the above description, the child process will send SIGCHLD signals to the parent process when its dies. by default, this signal is ignored by system, so the best way is that we can call wait() in the signal processing function, which could avoid the zombie stick around in the system. see more about this: http://itsprite.com/how-to-deep-understand-the-zombie-process-in-linux/

然后在编程时，如何摆脱僵尸进程的数量？根据以上描述，子进程在其死亡时将向父进程发送SIGCHLD信号。默认情况下，系统会忽略此信号，因此最好的方法是我们可以在信号处理函数中调用wait（），这样可以避免僵尸在系统中停留。了解更多相关信息：http：//itsprite.com/how-to-deep-understand-the-zombie-process-in-linux/

#1