分叉的最后一个孩子不会死-Lastchildforkedwillnotdie

I have the main process forking two times and thus creating two children. The two children are piped with each other like this:

我有两次主要的过程,因此创造了两个孩子。这两个孩子彼此用管道输送:

ls | more

Now the problem is that the second child never dies. Why is that? When does the last child in a pipe die really?

现在问题是第二个孩子永远不会死。这是为什么?管道中的最后一个孩子什么时候真的死了?

Removing one wait() call shows the expected result of ls | more but gives some further weird behaviours(stuck terminal etc).

删除一个wait()调用会显示ls |的预期结果更多,但给出了一些更奇怪的行为(卡住终端等)。

Here is my code:

这是我的代码:

int main(){
  printf("[%d] main\n", getpid());
  int pip[2], i;
  pipe(pip);

  /* CHILDREN*/
  for (i=0; i<2; i++){
    if (fork()==0){

      /* First child */
      if (i==0){
        printf("[%d] child1\n", getpid());
        close(1); dup(pip[1]);
        close(pip[0]);
        execlp("ls", "ls", NULL);}

      /* Second child */
      if (i==1){
        printf("[%d] child2\n", getpid());
        close(0); dup(pip[0]);
        close(pip[1]);
        execlp("more", "more", NULL);}
    }  
  }
  wait(NULL);  // wait for first child
  wait(NULL);  // wait for second child
  return 0;
}

2 个解决方案