Subject: Re: [Patch net] rds: mark bound socket with SOCK_RCU_FREE

From: Sowmini Varadhan <[email-masked]>
Date: Mon, 10 Sep 2018 19:30:07 -0400

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