问题描述
我正在编写自己的内核模块，它捕获 vfs_mkdir（struct inode *，struct dentry *，int）内核函数调用，并尝试记录磁盘路径名
我想用 dentry_path 内核函数来转换 struct dentry * 到一个路径名。
当我插入模块时，出现错误
我的内核版本是2.6.32，它应该被导出。
我找不出原因。
是否有其他的选择？
使用。 dentry_path 未被导出。
从linux-fsdevel ：
在星期五，四月20， 2012 at 02:08:37 PM -0400，Theodore Ts'o写道：
>我想知道如果我们简单地导出dentry_path（），
>也许是EXPORT_SYMBOL_GPL，并且警告它只能用于
>出于调试的目的，或者其他的一些。我怀疑这是不值得改变
>所有的inode_ops接口都要传递一个
>的结构路径intead。结构dentry，如果它只是用于调试。或者，也许我应该
>只要继续做这些丑陋的事情，并证明它们，因为它只是
>用于调试（yelch）。
>
>你怎么看？
只要使用dentry_path_raw（） - 它是被导出的，唯一的区别是
没有被删除。
I'm writing my own kernel module which captures vfs_mkdir(struct inode *, struct dentry *, int) kernel function invocation and tries to log the on-disk pathname where this invocation occurs.
I want to use the dentry_path kernel function to convert struct dentry * to a pathname.It's wired that when I insert the module, I get an error
My kernel version is 2.6.32 and it is supposed to be exported.I can't figure out the reason.Is there any alternatives?
Use dentry_path_raw. dentry_path isn't exported.
From linux-fsdevel archives:
On Fri, Apr 20, 2012 at 02:08:37PM -0400, Theodore Ts'o wrote:
> I wonder if we would be better off simply exporting dentry_path(),
> perhaps as EXPORT_SYMBOL_GPL, with a warning that it should only be used
> for debugging purposes, or some such. I suspect it's not worth changing
> all of the inode_ops interfaces to pass in a struct path intead of a
> struct dentry if it's only to be used for debugging. Or maybe I should
> just keep on doing these ugly things and justify them because it's only
> for debugging (yelch).
>
> What do you think?
Just use dentry_path_raw() - it _is_ exported and the only difference is
the lack of //deleted for unlinked ones.
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