如何从内核模块内的文件描述符获取文件名? [英] How can I get a filename from a file descriptor inside a kernel module?

问题描述

我需要从给定的文件描述符中获取一个文件的名称，在我编写的一个小的 linux 内核模块中.我尝试了 Getting Filename from file descriptor in C 中给出的解决方案，但出于某种原因，它会打印出垃圾值(如解决方案中所述，在 /proc/self/fd/NNN 上使用 readlink 时).我该怎么做?

I need to get the name of a file from a given file descriptor, inside a small linux kernel module that I wrote. I tried the solution given at Getting Filename from file descriptor in C, but for some reason, it prints out garbage values (on using readlink on /proc/self/fd/NNN as mentioned in the solution). How can I do it?

推荐答案

Don't call SYS_readlink - 使用与 procfs 相同的方法，当这些链接之一是读.从fs/proc/base.c中的proc_pid_readlink()和proc_fd_link()中的代码开始.

Don't call SYS_readlink - use the same method that procfs does when one of those links is read. Start with the code in proc_pid_readlink() and proc_fd_link() in fs/proc/base.c.

总的来说，给定一个 int fd 和一个来自你感兴趣的任务(你已经参考过)的 struct files_struct *files，你想要做:

Broadly, given an int fd and a struct files_struct *files from the task you're interested in (which you have taken a reference to), you want to do:

char *tmp;
char *pathname;
struct file *file;
struct path *path;

spin_lock(&files->file_lock);
file = fcheck_files(files, fd);
if (!file) {
    spin_unlock(&files->file_lock);
    return -ENOENT;
}

path = &file->f_path;
path_get(path);
spin_unlock(&files->file_lock);

tmp = (char *)__get_free_page(GFP_KERNEL);

if (!tmp) {
    path_put(path);
    return -ENOMEM;
}

pathname = d_path(path, tmp, PAGE_SIZE);
path_put(path);

if (IS_ERR(pathname)) {
    free_page((unsigned long)tmp);
    return PTR_ERR(pathname);
}

/* do something here with pathname */

free_page((unsigned long)tmp);

如果您的代码在进程上下文中运行(例如，通过系统调用调用)并且文件描述符来自当前进程，那么您可以将 current->files 用于当前任务的struct files_struct *.

If your code is running in process-context (eg. invoked through a syscall) and the file descriptor is from the current process, then you can use current->files for the current task's struct files_struct *.

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