Linux 如何从内核模块内的文件描述符获取文件名？
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How can I get a filename from a file descriptor inside a kernel module?
提问by Siddhant
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 readlinkon /proc/self/fd/NNNas mentioned in the solution). How can I do it?
我需要从给定的文件描述符中获取文件名，在我编写的一个小的 linux 内核模块中。我尝试了从 C 中的文件描述符获取文件名中给出的解决方案，但由于某种原因，它打印出垃圾值（在解决方案中提到的使用readlinkon 时/proc/self/fd/NNN）。我该怎么做？
采纳答案by caf
Don't call SYS_readlink- use the same method that procfsdoes when one of those links is read. Start with the code in proc_pid_readlink()and proc_fd_link()in fs/proc/base.c.
不要调用SYS_readlink- 使用与procfs读取这些链接之一时相同的方法。从 inproc_pid_readlink()和proc_fd_link()in 中的代码开始fs/proc/base.c。
Broadly, given an int fdand a struct files_struct *filesfrom the task you're interested in (which you have taken a reference to), you want to do:
从广义上讲，给定您感兴趣的任务中的int fda 和 a struct files_struct *files（您已经参考了该任务），您想要执行以下操作：
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);
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->filesfor the current task's struct files_struct *.
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，那么您可以使用current->files当前任务的struct files_struct *.
