子进程终止时了解SIGCHLD

Question

我无法理解以下程序的输出.我观察到在子进程返回后,父进程在wait()之前没有休眠3秒.如果SIGCHLD设置为默认处理程序,则它会休眠3秒,调用wait并按预期返回.到底发生了什么？

# include <unistd.h>
# include <sys/types.h>
# include <stdio.h>
# include <sys/wait.h>
# include <signal.h>

void handler(int sig) {
printf("Iam in handler ...\n");
}

main() {

int status;
pid_t pid;

struct sigaction act;
//act.sa_flags=SA_NOCLDSTOP;
act.sa_handler=handler;
sigaction(SIGCHLD,&act,NULL);

if(!fork()) {
printf("child process id is  %d\n",getpid());
 return 1;
}  

printf("xxx ...\n");
sleep(3);
pid = wait(&status);
printf("process terminated is %d\n",pid);

}

output::

xxx ...
child process id is  2445
Iam in handler ...
process terminated is 2445

Answer 1

从睡觉的男人():

sleep()使调用线程休眠直到秒秒或信号到达而不被忽略.

您的孩子终止会导致信号唤醒您.

返回值来自sleep():

如果请求的时间已经过去,则为零;如果呼叫被信号处理程序中断,则为剩余的秒数.

如果你想帮助你"完成"睡眠,可以使用.

unsigned sleep_time = 3;
...
while((sleep_time = sleep(sleep_time)) > 0) {}
pid = wait(&status);
...

Answer 2

当子进程死亡时,将a SIGCHLD发送给父进程.在你的情况下,它会中断sleep,看起来好像进程没有睡眠.

问题的要点:sleep当信号中断时不会重新启动.