linux-kernel - Re: unprivileged mounts git tree

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Message-Id: <E1KbJBO-0003Xz-0G@pomaz-ex.szeredi.hu>
Date:	Thu, 04 Sep 2008 20:03:50 +0200
From:	Miklos Szeredi <miklos@...redi.hu>
To:	serge@...lyn.com
CC:	miklos@...redi.hu, serue@...ibm.com, ebiederm@...ssion.com,
	akpm@...ux-foundation.org, hch@...radead.org,
	viro@...IV.linux.org.uk, linux-fsdevel@...r.kernel.org,
	linux-kernel@...r.kernel.org
Subject: Re: unprivileged mounts git tree

On Thu, 4 Sep 2008, Serge E. Hallyn wrote:
> Quoting Miklos Szeredi (miklos@...redi.hu):
> > On Thu, 4 Sep 2008, Serge E. Hallyn wrote:
> > > but you're still doing
> > > 
> > > 	if (IS_MNT_SHARED(old_nd.path.mnt) && !capable(CAP_SYS_ADMIN))
> > > 		goto out;
> > > 
> > > shouldn't it be something like
> > > 
> > > 	if (IS_MNT_SHARED(old_nd.path.mnt) && (old_nd.path.mnt & MNT_USER))
> > > 		goto out;
> > > 
> > > ?
> > 
> > Why would that be an error?  There's no real security gain to be had
> > from restricting a privileged user, but could cause a lot of
> > annoyance.  If we think this is dangerous, then protection should be
> > built into mount(8) with an option to override.  But not into the
> > kernel, IMO.
> 
> We disagree on that.  But can we agree that the check you added is wrong?

No :)

> There is no reason why a user mount should not be able to do shared
> mounts, is there?

I don't know.  It's something to think about in the future, but not
essential.  We know that without the above check the user can do bad
things: propagate mounts back into the source, and we don't want that.

We could allow binding a shared mount if

  a) the owners of the source and destination match
  b) the destination is made a slave of the source

But the current patchset doesn't allow _any_ changes to propagation
without CAP_SYS_ADMIN, so why should bind be an exception?

And yes, this is something to think about, but I think it's a rather
uncommon corner case, and so the patchset very much makes sense
without having to deal with unprivileged mount propagation changes.

>  So should the check above just go away then?

No, we'd be back with the original problem.

Thanks,
Miklos
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