从Linux内核模块的文件描述符中获取文件名/路径? [英] Get file name/path from a file descriptor from a Linux kernel module?

问题描述

在Linux内核模块中，是否有一种方法可以从unsigned int fd获取文件名/路径?

In a linux kernel module is there a way to get a file name/path from an unsigned int fd?

我知道这个答案:

I'm aware of this answer: How can I get a filename from a file descriptor inside a kernel module? but if I understand the code right, I need a struct files_struct too.

请停止重复而不是重复的投票.我正在寻求一种从内核模块中以纯C语言获取文件名/路径的方法，而不是使用系统工具. 换句话说:在/procself/fd/上运行readlink并不是很好的答案.

Please stop voting as duplicated as it isn't. I'm asking for a way to get file's name/path in plain C from a kernel module, not using system tools. Said in another way: running readlink on /procself/fd/ is not a good answer.

内核的系统调用读取ssize_t read(int fd, void *buf, size_t count);带有3个参数，其中之一是fd.显然，read能够以某种方式从单个文件(而不是inode内的所有文件)读取.问题是如何.

Kernel's syscall read ssize_t read(int fd, void *buf, size_t count); takes 3 arguments, one of them being a fd. It's obvious that somehow read is able to read from a single file (instead of all files inside an inode). The question is how.

推荐答案

您所引用问题的答案中的代码是您需要执行的操作.是的，需要来自任务的struct files_struct，因为文件描述符仅在files_struct的上下文中才有意义(通常每个进程都有一个).文件描述符不是全局唯一的，只是单个打开文件表中的索引.

The code in the answer to the question that you reference is what you need to do. And yes, a struct files_struct from a task is needed, because a file descriptor is only meaningful in the context of a files_struct (usually, there is one of these per process). File descriptors aren't globally unique, just an index within an individual open file table.

如果您的代码在进程上下文中运行(例如，通过syscall调用)，则可以将current->files用于当前任务的files_struct.这就是read()的作用.

If your code is running in process-context (eg. invoked through a syscall) then you can use current->files for the current task's files_struct. This is what read() does.

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