使用fork和dup的Unix管道

Question

可以说我的程序中我想执行两个子进程，一个执行“ls -al”命令，然后将其传送到“wc”命令并在终端上显示输出。我怎样才能做到这一点使用管道文件描述符到目前为止，我写的代码：一个例子是大大有益的

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

    pipe(pipefd2); 
    if ((fork()) == 0) { 
     dup2(pipefd2[1],STDOUT_FILENO); 
     close(pipefd2[0]); 
     close(pipefd2[1]); 
     execl("ls", "ls","-al", NULL); 
     exit(EXIT_FAILURE); 
    } 

    if ((fork()) == 0){ 
     dup2(pipefd2[0],STDIN_FILENO); 
     close(pipefd2[0]); 
     close(pipefd2[1]); 
     execl("/usr/bin/wc","wc",NULL); 
     exit(EXIT_FAILURE); 
    } 
    close(pipefd[0]); 
    close(pipefd[1]); 
    close(pipefd2[0]); 
    close(pipefd2[1]); 
} 

Answer 1

你的示例代码是语法和语义损坏（如pipefd2不decared，pipefd和pipefd2之间的混淆，等等）由于这种味道像家庭作业，请确保您了解我的注释，并询问更多您是否需要。我已经忽略了对pipe，fork和dup的错误检查，但理想情况下它们应该在那里。

int main(int argc, char *argv[]) { 
    int pipefd[2]; 
    pid_t ls_pid, wc_pid; 

    pipe(pipefd); 

    // this child is generating output to the pipe 
    // 
    if ((ls_pid = fork()) == 0) { 
     // attach stdout to the left side of pipe 
     // and inherit stdin and stdout from parent 
     dup2(pipefd[1],STDOUT_FILENO); 
     close(pipefd[0]);    // not using the right side 

     execl("/bin/ls", "ls","-al", NULL); 
     perror("exec ls failed"); 
     exit(EXIT_FAILURE); 
    } 

    // this child is consuming input from the pipe 
    // 
    if ((wc_pid = fork()) == 0) { 
     // attach stdin to the right side of pipe 
     // and inherit stdout and stderr from parent 
     dup2(pipefd[0], STDIN_FILENO); 

     close(pipefd[1]);    // not using the left side 
     execl("/usr/bin/wc", "wc", NULL); 
     perror("exec wc failed"); 
     exit(EXIT_FAILURE); 
    } 

    // explicitly not waiting for ls_pid here 
    // wc_pid isn't even my child, it belongs to ls_pid 

    return EXIT_SUCCESS; 
}