> On 08/02/14 16:45, Simon Wunderlich wrote:
> > Since batadv_orig_node_new() sets the refcount to two, assuming that
> > the calling function will use a reference for putting the orig_node into
> > a hash or similar, both references must be freed if initialization of
> > the orig_node fails. Otherwise that object may be leaked in that error
> > case.
> >
> > Reported-by: Antonio Quartulli <[email protected]>
> > Signed-off-by: Simon Wunderlich <[email protected]>
> > ---
> >
> > bat_iv_ogm.c | 2 ++
> > 1 file changed, 2 insertions(+)
> >
> > diff --git a/bat_iv_ogm.c b/bat_iv_ogm.c
> > index 6f4fcdc..6000337 100644
> > --- a/bat_iv_ogm.c
> > +++ b/bat_iv_ogm.c
> > @@ -256,6 +256,8 @@ batadv_iv_ogm_orig_get(struct batadv_priv *bat_priv,
> > const uint8_t *addr)
> >
> > free_bcast_own:
> > kfree(orig_node->bat_iv.bcast_own);
> >
> > free_orig_node:
> > + /* free twice, as batadv_orig_node_new set refcount to 2 */
> > + batadv_orig_node_free_ref(orig_node);
> >
> > batadv_orig_node_free_ref(orig_node);
>
> Coudln't we just invoke kfree(orig_node) here ? I think that if we hit
> this point it is because the node has not added to the hash and
> therefore it i snot used in any other context. This way we avoid the
> double free_ref() and we don't trgger the whole RCU mechanism.
>
> Or am I missing something?
What about already allocated substructures like bat_iv.bcast_own,
bat_iv.bacst_own_sum etc? Of course we could find out which is already
allocated and free that too, but that orig node free function does already
that.
Cheers,
Simon
