On 30/10/2024 17:37, Sabrina Dubroca wrote:
2024-10-29, 11:47:19 +0100, Antonio Quartulli wrote:
+static void ovpn_peer_release(struct ovpn_peer *peer) +{
- ovpn_bind_reset(peer, NULL);
- dst_cache_destroy(&peer->dst_cache);
Is it safe to destroy the cache at this time? In the same function, we use rcu to free the peer, but AFAICT the dst_cache will be freed immediately:
void dst_cache_destroy(struct dst_cache *dst_cache) { [...] free_percpu(dst_cache->cache); }
(probably no real issue because ovpn_udp_send_skb gets called while we hold a reference to the peer?)
Right. That was my assumption: release happens on refcnt = 0 only, therefore no field should be in use anymore. Anything that may still be in use will have its own refcounter.
- netdev_put(peer->ovpn->dev, &peer->ovpn->dev_tracker);
- kfree_rcu(peer, rcu);
+}
[...]
+static int ovpn_peer_del_p2p(struct ovpn_peer *peer,
enum ovpn_del_peer_reason reason)- __must_hold(&peer->ovpn->lock)
+{
- struct ovpn_peer *tmp;
- tmp = rcu_dereference_protected(peer->ovpn->peer,
lockdep_is_held(&peer->ovpn->lock));- if (tmp != peer) {
DEBUG_NET_WARN_ON_ONCE(1);if (tmp)ovpn_peer_put(tmp);Does peer->ovpn->peer need to be set to NULL here as well? Or is it going to survive this _put?
First of all consider that this is truly something that we don't expect to happen (hence the WARN_ON). If this is happening it's because we are trying to delete a peer that is not the one we are connected to (unexplainable scenario in p2p mode).
Still, should we hit this case (I truly can't see how), I'd say "leave everything as is - maybe this call was just a mistake".
Cheers,
return -ENOENT;- }
- tmp->delete_reason = reason;
- RCU_INIT_POINTER(peer->ovpn->peer, NULL);
- ovpn_peer_put(tmp);
- return 0;
+}