用shell中的管道连接n个命令？

Question

我试图在C中实现一个shell.我可以用一个简单的execvp()来执行简单的命令,但其中一个要求是管理这样的命令:"ls -l | head | tail -4"with a for '循环并且只有一个'pipe()'语句重定向stdin和stdout.几天后,我有点迷失了.

N =简单命令的数量(示例中为3:ls,head,tail)命令=带有命令的结构列表,如下所示:

commands[0].argv[0]: ls
commands[0].argv[1]: -l
commands[1].argv[0]: head
commands[2].argv[0]: tail
commands[2].argv[1]: -4

所以,我做了for循环,并开始重定向stdin和stdout以便用管道连接所有命令,但是...我只是无能为力它为什么不起作用.

for (i=0; i < n; i++){

pipe(pipe);
if(fork()==0){  // CHILD

    close(pipe[0]);
    close(1);
    dup(pipe[1]);
    close(pipe[1]);

    execvp(commands[i].argv[0], &commands[i].argv[0]);
    perror("ERROR: ");
    exit(-1);

}else{      // FATHER

    close(pipe[1]);
    close(0);
    dup(pipe[0]);
    close(pipe[0]);

}
}

我想要创建的是一系列childed进程:

[ls -l] ---- pipe ----> [head] ---- pipe ----> [tail -4]

所有这些进程都有一个root(运行我的shell的进程)所以,第一个父亲也是shell进程的一个孩子,我已经有点筋疲力尽了,有人可以帮助我吗？

我甚至不确定孩子是否应该是执行命令的孩子.

多谢你们 !!

Answer 1

这里没有什么复杂的,只要记住最后一个命令应该输出到原始进程'文件描述符1,第一个应该从原始进程文件描述符0读取.你只是按顺序生成进程,沿着输入端传输previois pipe电话.

所以,以下是类型:

#include <unistd.h>

struct command
{
  const char **argv;
};

使用简单明确定义的语义创建一个辅助函数:

int
spawn_proc (int in, int out, struct command *cmd)
{
  pid_t pid;

  if ((pid = fork ()) == 0)
    {
      if (in != 0)
        {
          dup2 (in, 0);
          close (in);
        }

      if (out != 1)
        {
          dup2 (out, 1);
          close (out);
        }

      return execvp (cmd->argv [0], (char * const *)cmd->argv);
    }

  return pid;
}

这是主叉例程:

int
fork_pipes (int n, struct command *cmd)
{
  int i;
  pid_t pid;
  int in, fd [2];

  /* The first process should get its input from the original file descriptor 0.  */
  in = 0;

  /* Note the loop bound, we spawn here all, but the last stage of the pipeline.  */
  for (i = 0; i < n - 1; ++i)
    {
      pipe (fd);

      /* f [1] is the write end of the pipe, we carry `in` from the prev iteration.  */
      spawn_proc (in, fd [1], cmd + i);

      /* No need for the write end of the pipe, the child will write here.  */
      close (fd [1]);

      /* Keep the read end of the pipe, the next child will read from there.  */
      in = fd [0];
    }

  /* Last stage of the pipeline - set stdin be the read end of the previous pipe
     and output to the original file descriptor 1. */  
  if (in != 0)
    dup2 (in, 0);

  /* Execute the last stage with the current process. */
  return execvp (cmd [i].argv [0], (char * const *)cmd [i].argv);
}

还有一个小测试:

int
main ()
{
  const char *ls[] = { "ls", "-l", 0 };
  const char *awk[] = { "awk", "{print $1}", 0 };
  const char *sort[] = { "sort", 0 };
  const char *uniq[] = { "uniq", 0 };

  struct command cmd [] = { {ls}, {awk}, {sort}, {uniq} };

  return fork_pipes (4, cmd);
}

似乎工作.:)