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 antonio@meshcoding.com Signed-off-by: Simon Wunderlich sw@simonwunderlich.de
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