C语言 经典 C. 在 execvp 函数中使用管道，stdin 和 stdout 重定向

Classic C. Using pipes in execvp function, stdin and stdout redirection

提问by krzakov

I want to simulate bash in my Linux C program using pipes and execvp function. e.g

我想在我的 Linux C 程序中使用管道和 execvp 函数模拟 bash。例如

ls -l | wc -l  

There is my program:

有我的程序：

if(pipe(des_p) == -1) {perror("Failed to create pipe");}

if(fork() == 0) {    //first fork
  close(1);          //closing stdout
  dup(des_p[1]);     //replacing stdout with pipe write 
  close(des_p[0]);   //closing pipe read
  close(des_p[1]);   //closing pipe write

  if(execvp(bash_args[0], bash_args)) // contains ls -l
    /* error checking */
}
else {
  if(fork() == 0) {  //creating 2nd child
    close(0);        //closing stdin
    dup(des_p[0]);   //replacing stdin with pipe read
    close(des_p[1]); //closing pipe write
    close(des_p[0]); //closing pipe read

    if(execvp(bash_args[another_place], bash_args)) //contains wc -l
      /* error checking */
  }

  close(des_p[0]);
  close(des_p[1]);
  wait(0);
  wait(0);
}

This code actually runs, but doesn't do the right thing. What's wrong with this code? That's not working and I don't have a clue why.

这段代码实际上运行了，但没有做正确的事情。这段代码有什么问题？这是行不通的，我不知道为什么。

回答by Nicholas Wilson

You need to close the pipe fds in the parent, or the child won't receive EOF, because the pipe's still open for writing in the parent. This would cause the second wait()to hang. Works for me:

您需要关闭父级中的管道 fds，否则子级将不会收到 EOF，因为管道仍处于打开状态以在父级中写入。这将导致第二个wait()挂起。对我有用：

#include <unistd.h>
#include <stdlib.h>


int main(int argc, char** argv)
{
        int des_p[2];
        if(pipe(des_p) == -1) {
          perror("Pipe failed");
          exit(1);
        }

        if(fork() == 0)            //first fork
        {
            close(STDOUT_FILENO);  //closing stdout
            dup(des_p[1]);         //replacing stdout with pipe write 
            close(des_p[0]);       //closing pipe read
            close(des_p[1]);

            const char* prog1[] = { "ls", "-l", 0};
            execvp(prog1[0], prog1);
            perror("execvp of ls failed");
            exit(1);
        }

        if(fork() == 0)            //creating 2nd child
        {
            close(STDIN_FILENO);   //closing stdin
            dup(des_p[0]);         //replacing stdin with pipe read
            close(des_p[1]);       //closing pipe write
            close(des_p[0]);

            const char* prog2[] = { "wc", "-l", 0};
            execvp(prog2[0], prog2);
            perror("execvp of wc failed");
            exit(1);
        }

        close(des_p[0]);
        close(des_p[1]);
        wait(0);
        wait(0);
        return 0;
}

回答by Art

Read up on what the waitfunction does. It will wait until one child process exists. You're waiting for the first child to exit before you start the second child. The first child probably won't exit until there's some process that reads from the other end of the pipe.

阅读该wait函数的作用。它将等待，直到存在一个子进程。在开始第二个孩子之前，您正在等待第一个孩子退出。第一个孩子可能不会退出，直到有一些进程从管道的另一端读取。