1

I'm trying to implement multiple piping using a tutorial I got from this website. I seem to get a bad file descriptor error after executing the function that takes care of multiple piping. When I'm duping for the first time it sends me this error. Here's the code:

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);
    }
}

Maybe there's a leakage somewhere or it can't find the descriptor. I don't seem to know where the problem is. What have I done wrong? Thanks.

Edited code:

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);
    }
4

2 回答 2

2

好的，首先，有点奇怪——你的嵌套与你的大括号不一致。if (j != 0)并且if(command->next)看起来像相同的“级别”，但实际的大括号讲述了一个不同的故事：

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);
        }
    }
}

请询问您的 IDE、编辑器，或indent(1)重新缩进您的代码以反映代码的实际语法，以免您被误导性的空格弄糊涂。

其次，我认为您在较早的迭代中更改了j+=2from a但并没有完全这样做 - 在第一次调用中，您正在使用，在下一次调用中您正在使用. 第一次迭代发生了什么？它被忽略。这是故意的吗？仅在下一次迭代中引用（通过）。任何东西都会引用倒数第二个条目吗？这也是故意的吗？j++pipefds[j-2]pipefds[j+1]j-1jj+=2 .. [j-2]pipefds[]

于 2011-12-08T09:52:13.133 回答
0

这是问题的答案。希望它可以帮助那里的人。我最终决定增加j2 (j+=2)。函数countPipes(char*)只是一个简单的函数来计算一个管道的(|)数量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);
}
于 2011-12-09T01:07:08.263 回答