c - 在 C（进程）中使用两个管道读取后无法退出循环

Question

我已经查看了这个以及这个堆栈溢出链接。

我无法理解关闭管道写入端的过程。在下面的代码中，我有 3 个进程，一个父进程，一个父进程的子进程，一个子进程的子进程。我正在尝试为命令模拟管道 - cat xxx | grep 28 | sort。我为此编写了一些代码，它基本上创建了排序、“抓取”/过滤器并打印我的输入，但它在最后挂起。我必须 ctrl + c 退出。我的代码有点乱，但如果你能帮我找出问题，那就太好了。

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <fcntl.h>
#include <sys/types.h>
#include <sys/stat.h>

/**
 * Executes the command "cat scores | grep Lakers".  In this quick-and-dirty
 * implementation the parent doesn't wait for the child to finish and
 * so the command prompt may reappear before the child terminates.
 *
 */

int main(int argc, char **argv)
{
  int pipefd[2];
  int pipefd2[2];
  int pid;

  char *cat_args[] = {"cat", "scores", NULL};
  char *grep_args[] = {"grep", "28", NULL};
  char *sort_args[] = {"sort", NULL};
  

  // make a pipe (fds go in pipefd[0] and pipefd[1])

  if (pipe(pipefd) != 0){
    return 1;
  }

   if (pipe(pipefd2) != 0){
    return 1;
  }

  pid = fork();

  if (pid < 0) 
    { 
    fprintf(stderr, "fork Failed" ); 
    exit(EXIT_FAILURE);
    } 
  else if (pid == 0)
    {     
    
    int pid2;
    pid2 = fork();
    
    if (pid2 < 0){
      fprintf(stderr, "fork Failed" ); 
      return 1;
    }
    else if (pid2 == 0){
      // replace standard input with input part of pipe
//       close(0);
//       close(pipefd[1]);
//       close(pipefd2[1]);
      dup2(pipefd2[0], 0);

      // close unused hald of pipe
      
      close(pipefd2[0]);
      close(pipefd[1]);
      close(pipefd2[1]);
      close(pipefd[0]);

      // execute grep

      execvp("sort", sort_args);
      close(pipefd[1]);
      close(pipefd2[1]);
      exit(0);
    }else{
      // replace standard input with input part of pipe
      
//       close(pipefd[1]);
//       close(pipefd2[1]);
      dup2(pipefd[0], 0);
      dup2(pipefd2[1], 1);

      // close unused hald of pipe
      close(pipefd[0]);
      close(pipefd2[1]);
      close(pipefd[1]);
      close(pipefd2[0]);

      // execute grep

      execvp("grep", grep_args);
      waitpid(pid2, NULL, 0);
      
     
      close(pipefd[1]);
      close(pipefd[0]);
      close(pipefd2[1]);
      close(pipefd2[0]);
      exit(0);
      waitpid(pid2, NULL, 0);
    }
    }
  else
    {
//       close(pipefd[1]);
//       close(pipefd2[1]);
      dup2(pipefd[1], 1);

      // close unused unput half of pipe
      
      close(pipefd[1]);
      close(pipefd[0]);
      close(pipefd2[1]);
      close(pipefd2[0]);
      // execute cat

      execvp("cat", cat_args);
    
    exit(0);
    waitpid(pid, NULL, 0);
    }
  
  close(pipefd[1]);
  close(pipefd[0]);
  close(pipefd2[1]);
  close(pipefd2[0]);

}

这是我得到的输出。不确定它是否相关，但如您所见，结果按团队名称排序。它只是不会终止。

Houston       44      28      .611
Indiana 45      28      .616
Oklahoma City   44      28      .611
Utah    44      28      .611
^C

标签： cpipeforkwaitpidmultiple-processes

Answer 1

调用execvp会用新的进程映像替换当前进程映像。如果没有发生错误，那么之后您的代码将永远不会到达任何行，因此您的close()和waitpid()函数调用是无用的。

编辑

这是您问题的功能齐全的代码。评论应该是不言自明的。请注意，命令执行顺序不同，我正在等待进程完成。

从空管道读取将阻塞，直到有一些数据要读取或发生错误，所以这不是唯一的解决方案。

    #include <stdio.h>
    #include <stdlib.h>
    #include <unistd.h>
    #include <fcntl.h>
    #include <sys/types.h>
    #include <sys/stat.h>
    #include <sys/wait.h>
    static void die (const char *msg) {
        perror (msg);
        exit (EXIT_FAILURE);
    }
    
    int main (int argc, char** argv) {
        int pipefd[2];
        int pid;
        char *cat_args[] = {"cat", "scores", NULL};
        char *grep_args[] = {"grep", "28", NULL};
        char *sort_args[] = {"sort", NULL};
    
        //make a pipe (file descriptor to read is pipefd[0], fd to write is pipefd[1])
        if (pipe (pipefd) < 0) 
            die ("creating a pipe failed");
        pid = fork();
        if (pid < 0)
            die ("fork");
        else if (pid == 0) {
        //child process
        int pipefd2[2]; //only visible to the affected processes
        if (pipe (pipefd2) < 0)
            die ("pipe");
        int pid2;
        pid2 = fork();
        if (pid2 < 0)
            die ("fork");
        else if (pid2 == 0) {
            //child of child will execute cat command
            close (pipefd2[0]); //we only need to write to the second pipe. close its reading end
    
            //first pipe is for communication between parent and grandparent only
            close (pipefd[0]);
            close (pipefd[1]);
    
            dup2 (pipefd2[1], STDOUT_FILENO); //write the output to the second pipe instead of the standard output
            close (pipefd2[1]); //close writing end of second pipe
    
            execvp("cat", cat_args); //execute cat command
            die ("execvp should never return");
    
        }
        else {
            //child process will execute the grep command
            close (pipefd2[1]); // we only need to read from the second pipe. close its writing end
            close(pipefd[0]); //we won't read from the first pipe
    
    
            waitpid (pid2, NULL, 0); //wait for cat command to finish
    
            dup2 (pipefd2[0], STDIN_FILENO); //read from the second pipe instead of the stdin
            close (pipefd2[0]); //child finished. close reading end of second pipe
    
            dup2 (pipefd[1], STDOUT_FILENO); //write the results of grep command to first pipe instead of standard output
            close (pipefd[1]); //we dup2 the output, close the writing end of first pipe
    
            execvp ("grep", grep_args);
            die ("execvp should never return");
    
        }
        } else {
        //parent process will execute the sort command
        close (pipefd[1]); //we won't write to the first pipe
    
    
        waitpid (pid, NULL, 0); //wait for child to write grep output to the first pipe
    
        dup2 (pipefd[0], STDIN_FILENO); //read from the first pipe instead of stdin
        close (pipefd[0]); //child finished. close reading end of first pipe
    
        execvp ("sort", sort_args); //execute command
        die ("execvp should never return");
        }
        //exit (EXIT_SUCCESS); we don't need this. the programm will never reach this line
    }