写入标准输入和读取标准输出(UNIX/LINUX/C 编程) [英] Writing to stdin and reading from stdout (UNIX/LINUX/C Programming)

问题描述

我正在做一项作业，其中程序将文件描述符作为参数(通常来自 exec 调用中的父级)并从文件中读取并写入文件描述符，在我的测试中，我意识到如果我使用 0、1 或 2 作为文件描述符，程序将在命令行中工作并且不会给出错误.这对我来说很有意义，只是我可以写入 stdin 并将其显示在屏幕上.

I was working on an assignment where a program took a file descriptor as an argument (generally from the parent in an exec call) and read from a file and wrote to a file descriptor, and in my testing, I realized that the program would work from the command-line and not give an error if I used 0, 1 or 2 as the file descriptor. That made sense to me except that I could write to stdin and have it show on the screen.

对此有解释吗?我一直认为对标准输入/标准输出有一些保护，你当然不能从标准输出 fprintf 到标准输入或 fgets.

Is there an explanation for this? I always thought there was some protection on stdin/stdout and you certainly can't fprintf to stdin or fgets from stdout.

#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
int main()
{
    char message[20];
    read(STDOUT_FILENO, message, 20);
    write(STDIN_FILENO, message, 20);

    return 0;
}

推荐答案

尝试写入标记为只读或反之亦然的文件将导致 write 和 read 返回-1，失败.在这种特定情况下，stdin 和 stdout 实际上是同一个文件.本质上，在您的程序执行之前(如果您不进行任何重定向)，shell 会运行:

Attempting to write on a file marked readonly or vice-versa would cause write and read to return -1, and fail. In this specific case, stdin and stdout are actually the same file. In essence, before your program executes (if you don't do any redirection) the shell goes:

  if(!fork()){
       <close all fd's>
       int fd = open("/dev/tty1", O_RDWR);
       dup(fd);
       dup(fd);
       execvp("name", argv);
  }

因此，stdin、out 和 err 都是同一个文件描述符的副本，打开用于读取和写入.

So, stdin, out, and err are all duplicates of the same file descriptor, opened for reading and writing.

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