Re: i_uid_read()/i_uid_write() and friends

From: Eric W. Biederman
Date: Sat Apr 11 2015 - 00:08:32 EST

Anton Altaparmakov <anton@xxxxxxxxxx> writes:

> Hi,
> Is it intended that non-gpl file systems cannot use functions like
> i_uid_read() and i_uid_write() (introduced by Eric Biederman in 3.5
> kernel)?
> They resolve to the below (in include/linux/fs.h):
> static inline uid_t i_uid_read(const struct inode *inode)
> {
> return from_kuid(&init_user_ns, inode->i_uid);
> }
> static inline void i_uid_write(struct inode *inode, uid_t uid)
> {
> inode->i_uid = make_kuid(&init_user_ns, uid);
> }
> And both from_kuid() and make_kuid() are EXPORT_SYMBOL() so they are
> fine but the problem is that init_user_ns is EXPORT_SYMBOL_GPL() and
> because i_uid_read() and i_uid_write() are static inline it causes
> them to be unusable from non-gpl kernel modules...
> Same thing applies to i_gid_read() and i_gid_write().
> These seem pretty fundamental calls that a non-gpl file system should
> be able to call, no?

I believe you are asking should a deriviative work of a GPL'd piece of
software be licenseable under something other than the GPL. I don't see
any provision in the GPLv2 for such a thing.

I am pretty certain that kuid_t is unique to the linux kernel and any
use of it makes your software a derivative work.

All EXPORT_SYMBOL_GPL does is provide a hint that the author could not
imagine that you could possibly use it in something that was not a
derivative work. EXPORT_SYMBOL makes no comment whatsoever.

As long as you are complying with the license of the kernel I don't see
how the question of a non-gpl file system can make any sense.

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