Zombie process 是已经执行完毕,但在进程表中仍有一个条目的进程(父进程还没有读取它的退出代码,或者换句话说,它还没有被“收割”)。
孤儿进程是其父进程已结束的进程,尽管它仍在运行(其父进程已“过世”但仍“活着”)。在这种情况下,init
将采用它并等待它。
考虑一下:
int main(int argv, char *argc[]) {
pid_t p=fork();
if (p<0) {
perror("fork");
}
// child
if (p==0) {
exit(2);
}
// parent sleeps for 2 seconds
sleep(2);
return 1;
}
这里创建的子进程将成为僵尸 2 秒,但是当父进程结束时它的状态是什么?孤儿僵尸?
它在进程表中的条目会发生什么变化?
“孤儿僵尸”(如上)是否也被 init
收养并被收割?
最佳答案
根据 man 2 wait
:
A child that terminates, but has not been waited for becomes a "zombie". The kernel maintains a minimal set of information about the zombie process (PID, termination status, resource usage information) in order to allow the parent to later perform a wait to obtain information about the child. As long as a zombie is not removed from the system via a wait, it will consume a slot in the kernel process table, and if this table fills, it will not be possible to create further processes. If a parent process terminates, then its "zombie" children (if any) are adopted by init(8), which automatically performs a wait to remove the zombies.
当父进程结束时,子进程(即使是僵尸进程)将被init
收养。然后,正如您所说,init
将wait()
以获取其退出状态。
所以,我不认为“孤儿僵尸”是什么特例。
关于c - 僵尸进程在其父进程死亡后会去哪里?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24346126/