获取内核代码中的当前工作目录 [英] Getting current working directory within kernel code

问题描述

我正在处理一个项目，我需要知道目前可调用系统调用的可执行文件的工作目录。我认为这样的一些系统调用，如 open 将会使用这些信息。

你能告诉我如何获取字符串中当前的工作目录路径？

解决方案

您可以看看 getcwd 系统调用是如何实现的，看看如何做到这一点。

该系统调用在 fs / dcache.c 并调用：

  get_fs_root_and_pwd（current-> fs，& root，& pwd）; 
  

根和 pwd 是结构路径变量，

该函数定义为内联函数 include / linux / fs_struct.h ，其中还包含：

  static inline void get_fs_pwd（struct fs_struct * fs，struct path * pwd）
  

这似乎是你之后。

I am working on a project in which I need to know the current working directory of the executable which called the system call. I think it would be possible as some system calls like open would make use of that information.

Could you please tell how I can get the current working directory path in a string?

解决方案

You can look at how the getcwd syscall is implemented to see how to do that.

That syscall is in fs/dcache.c and calls:

get_fs_root_and_pwd(current->fs, &root, &pwd);

root and pwd are struct path variables,

That function is defined as an inline function in include/linux/fs_struct.h, which also contains:

static inline void get_fs_pwd(struct fs_struct *fs, struct path *pwd)

and that seems to be what you are after.

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