1

我想在我的程序中设置 2 个管道。我有 1 个管道工作正常，但我不知道在哪里放置第二个管道。

我的设置的伪代码如下所示，

这是带花括号的，对此感到抱歉

//the first pipe:   
pipe(pipe1) 

//the second pipe:  
pipe(pipe2) 

pid = fork()

if(pid == 0) {

  dup2(pipe1[1], 1)
  close(pipe1[0])
  execvp(beforepipe)
  }  

if(pid > 0) { //everything below is in here

  pid2 = fork()

  if(pid2 == 0){

    //pipe1
    dup2(pipe1[0],0)
    dup2(out,1)
    close(pipe1[1])
    execvp(afterpipe)

    //pipe2 does not work might need to be placed in different area
    dup2(pipe1[1],1)
    close(pipe1[0])
    execvp(beforepipe1)
    }

  if(pid2 > 0){
    close(pipe[0])
    close(pipe[1])
    wait() //this is an infinite for loop

    pid3 = fork()

    if(pid3 == 0){
      dup2(pipe2[0],0)
      dup2(out,1)
      close(pipe2[1])
      execvp(afterpipe2)
      }

    if(pid3 > 0) {
      close(pipe2[0])
      close(pipe2[1])
      wait()
      }
    }

第二个管道的位置错误或代码完全错误。

有什么建议么？

4

2 回答 2

0 
execvp(afterpipe)
//pipe2 does not work might need to be placed in different area
dup2(pipe1[1],1)
close(pipe1[0])
execvp(beforepipe1)

我认为 execvp() 没有返回。所以 execvp() 下面的代码是无关紧要的。

于 2013-04-22T03:55:35.387 回答
0

您的主要问题是您没有关闭足够多的文件描述符。给定input1当前目录中包含您的字符串“ eschew obfuscation\”的文件，此代码适用于我（但请注意必须关闭多少文件描述符！）。

  • 经验法则：如果管道是标准输入或输出的dup2()d 或dup()d，则关闭两个文件管道文件描述符。

示例代码（带有调试跟踪）：

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

/* command pipeline: cat input1 | tr a-z A-Z | tr \\ q */

int main(void)
{
    int pipe1[2];
    int pipe2[2];
    pid_t pid1;
    char *cmd1[] = { "cat", "input1",        0 };
    char *cmd2[] = { "tr",  "a-z",    "A-Z", 0 };
    char *cmd3[] = { "tr",  "\\",     "q",   0 };

    if (pipe(pipe1) != 0 || pipe(pipe2) != 0)
    {
        perror("pipe failed");
        return 1;
    }

    pid1 = fork();

    if (pid1 < 0)
    {
        perror("fork 1 failed");
        return 1;
    }

    if (pid1 == 0)
    {
        /* Child 1 - cat */
        dup2(pipe1[1], 1);
        close(pipe1[0]);
        close(pipe1[1]);
        close(pipe2[0]);
        close(pipe2[1]);
        execvp(cmd1[0], cmd1);
        perror("failed to execute cmd1");
        return 1;
    }

    printf("pid 1 = %d\n", pid1);
    fflush(stdout);

    pid_t pid2 = fork();
    if (pid2 < 0)
    {
        perror("fork 2 failed");
        return 1;
    }

    if (pid2 == 0)
    {
        /* Child 2 - tr a-z A-Z */
        dup2(pipe1[0], 0);
        dup2(pipe2[1], 1);
        close(pipe1[0]);
        close(pipe1[1]);
        close(pipe2[0]);
        close(pipe2[1]);
        execvp(cmd2[0], cmd2);
        perror("failed to execute cmd2");
        return 1;
    }

    printf("pid 2 = %d\n", pid2);
    fflush(stdout);

    pid_t pid3 = fork();
    if (pid3 < 0)
    {
        perror("fork 3 failed");
        return 1;
    }

    if (pid3 == 0)
    {
        /* Child 3 - tr \\ q */
        dup2(pipe2[0], 0);
        close(pipe1[0]);
        close(pipe1[1]);
        close(pipe2[0]);
        close(pipe2[1]);
        execvp(cmd3[0], cmd3);
        perror("failed to execute cmd3");
        return 1;
    }

    printf("pid 3 = %d\n", pid3);
    fflush(stdout);

    /* Parent - wait for the kids to all die */
    close(pipe1[0]);
    close(pipe1[1]);
    close(pipe2[0]);
    close(pipe2[1]);

    pid_t corpse;
    int   status;
    while ((corpse = wait(&status)) > 0)
        printf("Child %d died status 0x%.4X\n", corpse, status);

    return 0;
}
于 2013-04-22T21:46:48.827 回答