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SubjectRe: [PATCH] VFS: fix recent breakage of FS_REVAL_DOT
On Mon, 2010-05-24 at 17:47 +0100, Al Viro wrote: 
> On Mon, May 24, 2010 at 12:21:22PM -0400, Trond Myklebust wrote:
> > > Can an nfs4 server e.g. have /x/y being a symlink that resolves to /a/b and
> > > allow mounting of both /x/y/c and /a/b/c? Which path would it return to
> > > client that has mounted both, walked to some referral point and called
> > > nfs_do_refmount(), triggering nfs4_proc_fs_locations()?
> > >
> > > Trond, Neil?
> >
> > When mounting /x/y/c in your example above, the NFSv4 protocol requires
> > the client itself to resolve the symlink, and then walk down /a/b/c
> > (looking up component by component), so it will in practice not see
> > anything other than /a/b/c.
> >
> > If it walks down to a referral, and then calls nfs_do_refmount, it will
> > do the same thing: obtain a path /e/f/g on the new server, and then walk
> > down that component by component while resolving any symlinks and/or
> > referrals that it crosses in the process.
> Ho-hum... What happens if the same fs is mounted twice on server? I.e.
> have ext2 from /dev/sda1 mounted on /a and /b on server, then on the client
> do mount -t nfs foo:/a /tmp/a; mount -t nfs foo:/b /tmp/b. Which path
> would we get from GETATTR with fs_locations requested, if we do it for
> /tmp/a/x and /tmp/b/x resp.? Dentry will be the same, since fsid would
> match.
> Or would the server refuse to export things that way?

I believe that the answer is that most filehandle types include an
encoding of the inode number of the export directory. In other words, as
long as '/a' and '/b' are different directories, then they will result
in the generation of different filehandles for /a/x and /b/x.

It seems that is not always the case, though. According to the
definition of mk_fsid(), it looks as if the 'FSID_UUID8' and
'FSID_UUID16' filehandle types only encode the uuid of the filesystem,
and have no inode information. They will therefore not be able to
distinguish between an export through '/a' or '/b'.

Neil, Bruce am I right?


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