Question

当我使用 ls\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\



int unipipe(char* lhs[], char* rhs[])
{
    int pfd[2];
    int status, cid;
    pid_t pid;
    char buf;
    if((lhs != NULL) && (rhs != NULL))
    {
      if(pipe(pfd) != 0)
      {
        perror("pipe");
        return -1;
      }
    if((pid = fork()) < 0)
    {
        perror("fork");
        return -1;  
    }
    else if(pid == 0)
   {
       close(1); //close the unused read end
      dup2(pfd[1], STDOUT_FILENO);
      //execute the left-hand side command
      close(pfd[0]);
      execvp(lhs[0], lhs);
       _exit(EXIT_SUCCESS);
    }

   if(setpgid(pid, 0) < 0)
   {
     perror("setpgid"); 
     return -1;
   };

  cid = waitpid(pid, &status, 0);
  if((pid = fork()) == 0)
    {
        close(0);
        dup2(pfd[0], STDIN_FILENO);
        close(pfd[1]);  //close the unused write end
        execvp(rhs[0], rhs);
        _exit(EXIT_SUCCESS);
    } 
    else
    {
           waitpid(pid, &status, 0);
    }
}

Answer 1

您等待第一个进程在开始第二个进程之前退出。每个管道都有一个缓冲, 一旦这个缓冲满了, I/ O 函数区块, 等待从管道读取几字节, 以便有更多的“ 流入 ” 。机会是您的第一个进程在管道上被屏蔽, 因此永远无法退出。

d 我宣布了两种Pid_t型变数,每个孩子一个, 并且等到这两个变数都成功启动后才开始。

Answer 2

要让程序运行,请删除第一个:

cid = waitpid(pid, &status, 0);

时间 :

else
{
       waitpid(pid, &status, 0);
}

替换为:

wait(); // for the fist child
wait(); // for the second child

您的程序将会运行。

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