On Tue, 2018-07-03 at 08:46 -0700, Tejun Heo wrote:
> Hello,
>
> On Tue, Jul 03, 2018 at 03:22:47PM +1000, Benjamin Herrenschmidt wrote:
> > +bool kernfs_has_children(struct kernfs_node *kn)
> > +{
> > + bool has_children = false;
> > + struct kernfs_node *pos;
> > +
> > + /* Lockless shortcut */
> > + if (RB_EMPTY_NODE(&kn->rb))
> > + return false;
>
> Hmm... shouldn't it be testing !rb_first(kn->dir.children)? The above
> would test whether @kn itself is unlinked.
Ah possibly, the rb tree usage got me confused a few times in there.
> > +
> > + /* Now check for active children */
> > + mutex_lock(&kernfs_mutex);
> > + pos = NULL;
> > + while ((pos = kernfs_next_descendant_post(pos, kn)) && pos != kn) {
> > + if (kernfs_active(pos)) {
> > + has_children = true;
> > + break;
> > + }
> > + }
> > + mutex_unlock(&kernfs_mutex);
> > +
> > + return has_children;
>
> The active ref is there to synchronize removal against in-flight reads
> so that kernfs_remove() can wait for them to drain. On return from
> kernfs_remove(), the node is guaranteed to be off the rbtree and the
> above test isn't necessary. !rb_first() test should be enough
> (provided that there's external synchronization, of course).
Ok. Yes there's external synchronization. So a simple !rb_first test
doesn't need the mutex I suppose ?
Cheers,
Ben.
>
> Thanks.
>
