是否有Linux或POSIX方法用于指示进程在完成且父进程未调用waitpid()
时变为僵尸?
我知道父进程我们可以使用SA_NOCLDSTOP
来处理SIGCHLD
处理程序,但在我的情况下这不是一个选项,因为父进程繁忙且SIGCHLD
正在用于其他的东西。
如果我对退出代码不感兴趣,有没有办法将特定的子进程标记为自己悄然死亡?
答案 0 :(得分:0)
您将始终需要等待子进程,但是如果您按照此过程进行操作,那么您可以等待一个快速死亡的孩子,并让init继承您的真实'处理。然后,Init会为你整理。
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
void main(void) {
int ret;
pid_t child1;
pid_t child2;
int status;
child1 = fork();
if (child1 == -1) {
/* error */
exit(1);
}
if (child1 == 0) {
/* in the child... we create a new session, and then re-fork */
setsid();
child2 = fork();
if (child2 == -1) {
exit(1);
}
if (child2 == 0) {
/* call execve() or a friend */
ret = execlp("sleep", "sleep", "6", NULL);
/* we should _never_ get here - unless the execlp() fails */
fprintf(stderr, "execlp() returned: %d\n", ret);
exit(1);
for(;;);
}
sleep(2);
/* success ... child1 dies here */
exit(0);
}
sleep(4);
ret = waitpid(child1, &status, 0);
if (ret != 0) {
/* unfortunately we can only determine the state of our 'proxy' process...
* to get any further information / the child-child PID, then you'll need to use a pipe to share the information */
fprintf(stderr, "waitpid() returned %d\n", ret);
}
sleep(4);
return;
}
各种睡眠持续时间应允许您查看以下事件(观看top
或其他内容)。
所有进程都会启动,所有进程都作为shell的子进程链接
top
输出:
attie 17336 17335 0 16:04 pts/36 00:00:00 -bash
attie 21855 17336 0 16:34 pts/36 00:00:00 ./test
attie 21856 21855 0 16:34 ? 00:00:00 ./test
attie 21857 21856 0 16:34 ? 00:00:00 sleep 6
Child1 死亡,变成僵尸/不存在, Child2 被init(PID 1)继承
attie 17336 17335 0 16:04 pts/36 00:00:00 -bash
attie 21855 17336 0 16:34 pts/36 00:00:00 ./test
attie 21856 21855 0 16:34 ? 00:00:00 [test] <defunct>
attie 21857 1 0 16:34 ? 00:00:00 sleep 6
Child1 在调用waidpid()
attie 17336 17335 0 16:04 pts/36 00:00:00 -bash
attie 21855 17336 0 16:34 pts/36 00:00:00 ./test
attie 21857 1 0 16:34 ? 00:00:00 sleep 6
Child2 死亡,并被init清除
attie 17336 17335 0 16:04 pts/36 00:00:00 -bash
attie 21855 17336 0 16:34 pts/36 00:00:00 ./test