dup2 错误的文件描述符

Question

我正在尝试在我的 shell 中实现管道。现在它可以处理正常的命令，如 ls、who、date 等。

通过大量阅读并进入其他堆栈溢出帖子以检查管道应该如何工作，我想出了这段代码，我将向你们演示。该函数基本上接受一个应该处理的命令。

void
runCommand(Command *cmd){

    char **pl = cmd->pgm->pgmlist;
    int status;
    Pgm *p = cmd->pgm;

    //Count the number of pipes and create the right number of filedescriptors
    int numPipes = countPipes(cmd->pgm);
    int pipefds[2*numPipes];

    //Pipe the file descriptors here and check if there was an error.
    for(int i = 0; i < (numPipes); i++){
        if(pipe(pipefds + i*2) < 0) {
            perror("couldn't pipe");
            exit(EXIT_FAILURE);
        }
    }

    pid_t pid, wpid;


    int fdr;
    int fd;

    int j = 0;
    while(p != NULL) {

        pid = fork();

        if (pid == 0) {
            // Child process
            if (cmd->bakground == 1) // Check if it should be running in the background, if so, assign a new PID
            {
                setpgid(pid, 0);
            }

            // Check if RSTDOUT is on or not
            if(cmd->rstdout != NULL){

                fd = open(cmd->rstdout, O_WRONLY | O_CREAT | O_TRUNC , S_IRUSR | S_IRGRP | S_IWGRP | S_IWUSR);
                //fclose(fopen(cmd->rstdout, "w"));
                printf("in first\n");
                dup2(fd,1);
                close(fd);
            }

            // Check if RSTIN is on or not
            if(cmd->rstdin != NULL) {

                fdr = open(cmd->rstdin, O_RDONLY);
                printf("in second\n");
                dup2(fdr, 0);
                close(fdr);
            }

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

            //if not first command&& j!= 2*numPipes
            if(j != 0 ){

                printf("in fourth: %d\n", j);
                if(dup2(pipefds[j-2], 0) < 0){
                    perror(" dup2");///j-2 0 j+1 1
                    exit(EXIT_FAILURE);

                }

            }

            for(int i = 0; i < 2*numPipes; i++){
                    printf("in fifth\n");
                    close(pipefds[i]);
            }

            j += 2;
            printf("%s\n",pl[0] );
            if (execvp(pl[0], pl) == -1) {
                perror("lsh");
            }

            exit(EXIT_FAILURE);
        }

        else if (pid < 0) {
            // Error forking
            perror("lsh");
        } 
        else if(cmd->bakground != 1){ // check in the parent process if it was running in background, if so dont call for wait()
            // Parent process
            j+=2;

            pl = p->pgmlist;    
            p = p->next;
            for(int i = 0; i < 2 * numPipes; i++){
                printf("in sixth\n");
                close(pipefds[i]);
            }

            int returnStatus;    
            waitpid(pid, &returnStatus, 0);  // Parent process waits here for child to terminate.

            if (returnStatus == 0)  // Verify child process terminated without error.  
            {
               printf("The child process terminated normally.");    
            }

            if (returnStatus == 1)      
            {
               printf("The child process terminated with an error!.");    
            }

        } 
        else
            printf("Child process with ID %d is running in background\n",pid );
    }

}

普通命令仍然可以正常工作，但是当我尝试这样做时，我得到以下结果。

> ls | wc
>  dup2: Bad file descriptor

当我读到它时，我认为这意味着我只是接近早期，或者我处理我的文件描述符很糟糕。我试图修复它，但没有成功，我求助于你们，他们可能会看到一些我看不到的东西。如果您需要任何进一步的信息，请发表评论，我会尽力提供帮助。

******编辑 1******

我将 printf 添加到我使用的每个 dup2 或 close 中，将它们命名为 first、second、third 等，只是为了确切知道我在哪个 dup 或 close 中得到错误。这是我运行命令时的结果：

> ls | wc
in sixth
in sixth
in third
in sixth
in sixth
in fourth: 2
 dup2: Bad file descriptor

所以知道我们至少知道它失败的地方。

******编辑2******

我还添加了我们正在写入、读取或关闭的文件描述符：

In parent closing fd: 0
In parent closing fd: 1
in child third writing to fd: 1
In parent closing fd: 0
In parent closing fd: 1
in child fourth reading from fd: 0
 dup2: Bad file descriptor

Answer 1

我认为你的 dup2 问题就在这里：

if(cmd->rstdin != NULL) {

                fdr = open(cmd->rstdin, O_RDONLY);
                dup2(fd, 0);
                close(fdr);
            }

您正在打开“fdr”，但随后在 dup2 中使用了“fd”。 “fd”未打开或已在先前的 if {} 中关闭。此外，在这两个条件中，您都不会检查 dup2 的返回值。