Question

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

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?

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

Here is my code:

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;
}

Answer 1

The read end of the pipe won’t get an EOF mark until the write end has been closed by all its users. The parent of both children still has both ends of the pipe open, so more doesn’t see an EOF (a return of 0 from read()), and keeps waiting for more input.