On (09/10/18 15:43), Santosh Shilimkar wrote: > On 9/10/2018 3:24 PM, Cong Wang wrote: > >When a rds sock is bound, it is inserted into the bind_hash_table > >which is protected by RCU. But when releasing rd sock, after it > >is removed from this hash table, it is freed immediately without > >respecting RCU grace period. This could cause some use-after-free > >as reported by syzbot. > > > Indeed. > > >Mark the rds sock as SOCK_RCU_FREE before inserting it into the > >bind_hash_table, so that it would be always freed after a RCU grace > >period.
So I'm not sure I understand. Yes, Cong's fix may eliminate *some* of the syzbot failures, but the basic problem is not solved. To take one example of possible races (one that was discussed in https://www.spinics.net/lists/netdev/msg475074.html) rds_recv_incoming->rds_find_bound is being called in rds_send_worker context and the rds_find_bound code is 63 rs = rhashtable_lookup_fast(&bind_hash_table, &key, ht_parms); 64 if (rs && !sock_flag(rds_rs_to_sk(rs), SOCK_DEAD)) 65 rds_sock_addref(rs); 66 else 67 rs = NULL; 68 After we find an rs at line 63, how can we be sure that the entire logic of rds_release does not execute on another cpu, and free the rs, before we hit line 64 with the bad rs? Normally synchronize_rcu() or synchronize_net() in rds_release() would ensure this. How do we ensure this with SOCK_RCU_FREE (or is the intention to just reduce *some* of the syzbot failures)? --Sowmini