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我为我的班级分配了这个作业，这个程序几乎完成了，但输出与它应该是相去甚远。

我在 linux 系统上运行它并使用 geany ide。

这是该程序的主要要点：

实现一个 C++ 程序，提示用户输入 2 个命令。每个输入字符串应该是一个 UNIX 命令，允许使用参数。例如，输入 1 可以是“ls -l”，输入 2 可以是“wc -l”。然后程序将创建一个管道和两个子进程。第一个子进程将运行第一个输入中指定的命令。它将输出到管道而不是标准输出。第二个子进程将运行第二个输入中指定的命令。它将从管道而不是标准输入中获取其输入。父进程将等待其两个子进程完成，然后整个过程将重复。当第一个命令输入“quit”时，执行将停止。

到目前为止，这是我的代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <cstring>
using namespace std;

int main() {
    //Declare Variables
    char firstAns[80], secondAns[80];
    int rs, ansLen, pipefd[2];
    int pid1, pid2;
    char * command1[5], *command2[5];
    char quit[] = "quit\0";

    //Allocate memory for commands
    for(int a=0; a<5; a++){
        command1[a]=new char[80];
    }

    for(int b=0; b<5; b++){
        command2[b]=new char[80];
    }

    //Asking for the commands
    cout << "Please enter your first command(incl. args) or quit: ";
    cin >> firstAns;

    //Do the program while the first answer isn't quit
    while(firstAns != quit) {
        //Copy first answer into first command
        ansLen = strlen(firstAns);
        for(int i=0; i < ansLen; i++){
            strcpy(command1[i], firstAns);
        }

        //Just skip to end of program if first command is quit
        cout << "Please enter your second command(incl. args): ";
        cin >> secondAns;

        //Copy second answer into second command
        ansLen = strlen(secondAns);
        for(int i=0; i < ansLen; i++){
            strcpy(command1[i], secondAns);
        }

        //pipe
        rs = pipe(pipefd);
        //if pipe fails to be made
        if(rs == -1){
            perror("Fail to create a pipe");
            exit(EXIT_FAILURE);
        }

        //Fork for the two processes
        pid1 = fork();

        if (pid1 != 0){
            pid2 = fork();
        }

        if(pid1 == -1){
            perror("Fail to create a pipe");
            exit(EXIT_FAILURE);
        }
            if(pid2 == -1){
            perror("Fail to create a pipe");
            exit(EXIT_FAILURE);
        }

        if (pid1 == 0) { // 1st child process
            // close write end of pipe
            close(pipefd[0]);
            // duplicate
            dup2(pipefd[1], 1);
            //execute the input argument
            rs = execvp(command1[0], command1);
            //error check
            if (rs == -1) {
                perror("Fail to execute command");
                exit(EXIT_FAILURE);
            }
        }

        if (pid2 == 0) { // 2st child process
            // close write end of pipe
            close(pipefd[1]);
            // duplicate
            dup2(pipefd[0], 0);
            //execute the input argument
            rs = execvp(command2[0], command2);
            //error check
            if (rs == -1) {
                perror("Fail to execute command");
                exit(EXIT_FAILURE);
            }
        }
        else {// parent process
            // close read and write end of pipe
            close(pipefd[0]);
            close(pipefd[1]);
            // wait for child processes
            wait(&pid1);
            wait(&pid2);
        }

        //Asking for the commands
        cout << "Please enter your first command(incl. args) or quit: ";
        cin >> firstAns;

    }; //end of while()

    return 0;
}

我没有收到任何错误，但我没有得到我想要的输出。这是它应该如何工作的示例输出：

Enter command 1 (incl. args) or quit: ls
Enter command 2 (incl. args): wc
      5       5      63
Enter command 1 (incl. args) or quit: ls -l
Enter command 2 (incl. args): wc -w
      47
Enter command 1 (incl. args) or quit: hello
Enter command 2 (incl. args): more
hello: No such file or directory
Enter command 1 (incl. args) or quit: quit

相反，我的输出只产生 perror，然后说 No file or directory found for the first command and Bad address for the second command，如下所示：

Enter command 1 (incl. args) or quit: ls
Enter command 2 (incl. args): wc
Failed to execute command: No such file or directory
Failed to execute command: Bad address

我猜它是因为变量在 execpv 中不匹配，但我可以稍微推动一下最终击败这个程序。

如果有人对此也有任何建议，那么退出命令似乎也不起作用。任何类型的建议都会很棒，并且会帮助这个年轻的程序员:)

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