Question

我正在尝试使用我从website获得的教程来实现多个管道。在执行处理多个管道的功能后，我似乎得到了错误的文件描述符错误。当我第一次复制时，它会向我发送此错误。这是代码：

void runPipedCommands(cmdLine* command, char* userInput) {
    int numPipes = countPipes(userInput);

    int status = 0;
    int i = 0, j = 0;

    pid_t pid;

    int pipefds[2*numPipes];

    for(i = 0; i < (numPipes); i++){
        if(pipe(pipefds + i*2) < 0) {
            perror("pipe");
            exit(EXIT_FAILURE);
        }
    }

    j = 0;
    while(command) {
        pid = fork();
        if(pid == 0) {

            //if not first command
            if(j != 0){
                if(dup2(pipefds[j-2], 0) < 0){
                    perror(" dup2");///j-2 0 j+1 1
                    exit(EXIT_FAILURE);

                }

            if(command->next){
                printf(
                if(dup2(pipefds[j + 1], 1) < 0){
                    perror("dup2");
                    exit(EXIT_FAILURE);
                }
            }

            for(i = 0; i < 2*numPipes; i++){
                    close(pipefds[i]);
            }

            if( execvp(*command->arguments, command->arguments) < 0 ){
                    perror(*command->arguments);
                    exit(EXIT_FAILURE);
            }
        } else if(pid < 0){
            perror("error");
            exit(EXIT_FAILURE);
        }

        command = command->next;
        j++;
    }
        for(i = 0; i < 2 * numPipes; i++){
            close(pipefds[i]);
            printf("in parent: closed pipe[%d]\n", i);
        }
        wait(0);
    }
}

也许某处有泄漏或无法找到描述符。我似乎不知道问题出在哪里。我做错了什么？感谢。

编辑代码：

void runPipedCommands(cmdLine* command, char* userInput) {
    int numPipes = countPipes(userInput);

    int status = 0;
    int i = 0, j = 0;

    pid_t pid;

    int pipefds[2*numPipes];

    for(i = 0; i < (numPipes); i++){
        if(pipe(pipefds + i*2) < 0) {
            perror("pipe");
            exit(EXIT_FAILURE);
        }
    }

    j = 0;
    while(command) {
        pid = fork();
        if(pid == 0) {
            //if not first command
            if(j != 0 && j!= 2*numPipes){
                if(dup2(pipefds[j-2], 0) < 0){
                    perror(" dup2");///j-2 0 j+1 1
                    exit(EXIT_FAILURE);

                }
            }

            //if not last command
            if(command->next){
                printf("command exists: dup(pipefd[%d], 1])\n", j+1);
                if(dup2(pipefds[j + 1], 1) < 0){
                    perror("dup2");
                    exit(EXIT_FAILURE);
                }
            }

            for(i = 0; i < 2*numPipes; i++){
                    close(pipefds[i]);
                   printf("in child: closed pipe[%d]\n", i);
            }

            if( execvp(*command->arguments, command->arguments) < 0 ){
                    perror(*command->arguments);
                    exit(EXIT_FAILURE);
            }
        } else if(pid < 0){
            perror("error");
            exit(EXIT_FAILURE);
        }

        command = command->next;
        j+=2;
    }
        for(i = 0; i < 2 * numPipes; i++){
            close(pipefds[i]);
            printf("in parent: closed pipe[%d]\n", i);
        }   
           wait(0);
    }

Answer 1

好的，首先，有些奇怪的东西 - 你的嵌套不符合你的牙箍。 if (j != 0)和if(command->next)看起来像“级别”，但实际大括号会说明不同的故事：

复制和粘贴：

        if(j != 0){
            if(dup2(pipefds[j-2], 0) < 0){
                perror(" dup2");///j-2 0 j+1 1
                exit(EXIT_FAILURE);
            }

        if(command->next){
            printf(
            if(dup2(pipefds[j + 1], 1) < 0){
                perror("dup2");
                exit(EXIT_FAILURE);
            }
        }

重新缩进以反映大括号：

if (j != 0) {
    if (dup2(pipefds[j - 2], 0) < 0) {
        perror(" dup2");    ///j-2 0 j+1 1
        exit(EXIT_FAILURE);
    }

    if (command->next) {
        printf(); /* fixed this */
        if (dup2(pipefds[j + 1], 1) < 0) {
            perror("dup2");
            exit(EXIT_FAILURE);
        }
    }
}

请让您的IDE，编辑或indent(1)重新缩进代码以反映代码的实际语法，这样您就不会误导误导空格。

其次，我认为你在之前的迭代中从j+=2更改了j++但没有完全改变 - 在第一次调用中，你正在使用pipefds[j-2]和在下一个电话中，您正在使用pipefds[j+1]。无论第一次迭代中j-1发生了什么？它被忽略了。这是故意的吗？ j仅在 next 迭代中引用（通过j+=2 .. [j-2]）。有什么东西可以引用pipefds[]中的倒数第二个条目吗？这是故意的吗？

Answer 2

这是问题的答案。希望它可以帮助那里的人。我最终决定将j增加2 (j+=2)。函数countPipes(char*)只是一个简单的函数，用于计算来自(|)

的管道数char*

void runPipedCommands(cmdLine* command, char* userInput) {
    int numPipes = countPipes(userInput);

    int status;
    int i = 0;
    pid_t pid;

    int pipefds[2*numPipes];//declare pipes

    /**Set up pipes*/
    for(i = 0; i < (numPipes); i++){
        if(pipe(pipefds + i*2) < 0) {
            perror("couldn't pipe");
            exit(EXIT_FAILURE);
        }
    }

    int j = 0;
    while(command) {
        pid = fork();
        if(pid == 0) {

            //if not last command
            if(command->next){
                if(dup2(pipefds[j + 1], 1) < 0){
                    perror("dup2");
                    exit(EXIT_FAILURE);
                }
            }

            //if not first command
            if(j != 0 ){
                if(dup2(pipefds[j-2], 0) < 0){
                    perror(" dup2");///j-2 0 j+1 1
                    exit(EXIT_FAILURE);

                }
            }

            //close pipes in child
            for(i = 0; i < 2*numPipes; i++){
                    close(pipefds[i]);
            }

            //execute commands
            if( execvp(*command->arguments, command->arguments) < 0 ){
                    perror(*command->arguments);
                    exit(EXIT_FAILURE);
            }
        } else if(pid < 0){
            perror("error");
            exit(EXIT_FAILURE);
        }

        command = command->next;//go to the next command in the linked list
        j+=2;//increment j
    }

    /**Parent closes the pipes and waits for all of its children*/

    for(i = 0; i < 2 * numPipes; i++){
        close(pipefds[i]);
    }

    for(i = 0; i < numPipes + 1; i++) //parent waits for all of its children
        wait(&status);
}

dup2错误的文件描述符错误

2 个答案: