C语言 如何防止僵尸子进程？
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How can I prevent zombie child processes?
提问by thejh
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fork()用于为客户端连接生成处理程序的服务器。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，它们应该直接死亡而不是变成僵尸。有什么简单的方法可以实现这一目标？
采纳答案by thejh
回答by xaxxon
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 fork 另一个副本并退出原始子进程。
http://thinkiii.blogspot.com/2009/12/double-fork-to-avoid-zombie-process.html
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，也更容易理解。
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);
}
}
回答by user3060336
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.
您不能像杀死正常进程一样使用 SIGKILL 信号杀死僵尸进程，因为僵尸进程无法接收任何信号。所以养成一个好习惯很重要。
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/
那么在编程时，如何摆脱僵尸进程的数量呢？根据上面的描述，子进程死亡时会向父进程发送SIGCHLD信号。默认情况下，这个信号被系统忽略，所以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徘徊。查看更多相关信息：http: //itsprite.com/how-to-deep-understand-the-zombie-process-in-linux/
