如何让父进程等待所有子进程完成? [英] How to make parent wait for all child processes to finish?

问题描述

我希望有人可以阐明如何让父进程等待 ALL 子进程完成，然后再继续分叉.我有要运行的清理代码，但子进程需要在这发生之前返回.

I'm hoping someone could shed some light on how to make the parent wait for ALL child processes to finish before continuing after the fork. I have cleanup code which I want to run but the child processes need to have returned before this can happen.

for (int id=0; id<n; id++) {
  if (fork()==0) {
    // Child
    exit(0);      
  } else {
    // Parent
    ...
  }
  ...
}

推荐答案

pid_t child_pid, wpid;
int status = 0;

//Father code (before child processes start)

for (int id=0; id<n; id++) {
    if ((child_pid = fork()) == 0) {
        //child code
        exit(0);
    }
}

while ((wpid = wait(&status)) > 0); // this way, the father waits for all the child processes 

//Father code (After all child processes end)

wait 等待a 子进程终止，并返回该子进程的 pid.出错时(例如，当没有子进程时)，返回 -1.所以，基本上，代码一直在等待子进程完成，直到 waiting 错误出现，然后你就知道它们都完成了.

wait waits for a child process to terminate, and returns that child process's pid. On error (eg when there are no child processes), -1 is returned. So, basically, the code keeps waiting for child processes to finish, until the waiting errors out, and then you know they are all finished.

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