在 C 中实现多个管道和命令执行 [关闭]

Question

我有一个作业要求我创建一个 shell，它执行由管道分隔的多个命令。这是我的代码 sn-p 用于派生子和执行命令：

    for (int i = 0; i < commands; i++) {

        place = commandStarts[i];

        if((pid = fork()) < 0) exit(1);
        else if (pid == 0) {
            if(i < pipe_counter && dup2(fds[j + 1], 1) < 0) exit(1);// If this is not the last remaining command

            if(j != 0 && dup2(fds[j-2], 0) < 0) exit(1);        // If this is not the first command

            for(int c = 0; c < 2*pipe_counter; c++) close(fds[c]);

            if(execvp(argv[place], argv+place)) exit(0);        // Execute and if it returns anything, exit!
        }
        j+=2;
    }

    for(int i = 0; i < 2 * pipe_counter; i++) close(fds[i]);

    while ((wait(&status)) != pid);                     // The only program that gets to this point is the 
                                        //     parent process, which should wait for completion

尽管它在简单的例子中似乎工作得很好，在一些更复杂的例子中，分级系统给了我这个提示：在显示提示之前，你应该等待管道链中的所有进程终止，不只是最后一个！

你能告诉我错在哪里吗？

Answer 1

好吧，我的错误是我只等待最后一个分叉的孩子终止。

这是等待每个子节点终止的正确代码 sn-p：

    for (int i = 0; i < commands; i++) {

        place = commandStarts[i];

        if((pid = fork()) < 0) exit(1);
        else if (pid == 0) {
            if(i < pipe_counter && dup2(fds[j + 1], 1) < 0) exit(1);// If this is not the last remaining command

            if(j != 0 && dup2(fds[j-2], 0) < 0) exit(1);            // If this is not the first command

            for(int c = 0; c < 2*pipe_counter; c++) close(fds[c]);

            if(execvp(argv[place], argv+place)) exit(1);            // Execute and if it returns anything, exit!
        }
        j+=2;
    }

    for(int i = 0; i < 2 * pipe_counter; i++) close(fds[i]);

    for(int i = 0; i < commands; i++) wait(&status);            // The only program that gets to this point is the 
                                                                //     parent process, which should wait for completion.

也非常感谢 wallyk 帮助我思考这个问题！