Quoting Pekka J Enberg ([email protected]):
> Hi Serge,
>
> (Thanks for looking at this. I appreciate the review!)
>
> On Mon, 17 Dec 2007, [email protected] wrote:
> > > struct vfsmount *mnt = nd->mnt;
> > > - struct dentry *dentry = __d_lookup(nd->dentry, name);
> > > + struct dentry *dentry;
> > >
> > > +again:
> > > + dentry = __d_lookup(nd->dentry, name);
> > > if (!dentry)
> > > goto need_lookup;
> > > +
> > > + if (dentry->d_inode && IS_REVOKE_LOCKED(dentry->d_inode)) {
> >
> > not sure whether this is a problem or not, but dentry->d_inode isn't
> > locked here, right? So nothing is keeping do_lookup() returning
> > with an inode which gets revoked between here and the return 0
> > a few lines down?
>
> I assume you mean S_REVOKE_LOCK and not ->i_mutex, right?
No I did mean the i_mutex since you take the i_mutex when you set
S_REVOKE_LOCK. So between that and the comment above do_lookup(),
I assumed you were trying to lock out concurrent do_lookups() returning
an inode whose revoke is starting at the same time.
But based on your next paragraph it sounds like I misunderstand your
locking.
> The caller is supposed to block open(2) with chmod(2)/chattr(2) so while
> revoke is in progress, you can get references to the _revoked inode_,
> which is fine (operations on it will fail with EBADFS). The
> ->i_revoke_wait bits are there to make sure that while we revoke, you
> can't get a _new reference_ to the inode until we're done.
And a new reference means through iget(), so if revoke starts
between the IS_REVOKE_LOCKED() check in do_lookup and its return,
it's ok bc we'll get a reference later on?
I'm a little confused but i'll keep looking.
thanks,
-serge
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