On 17/12/18 - 14:01:41, Eric Dumazet wrote:
> On Mon, Dec 17, 2018 at 1:57 PM Christoph Paasch <cpaa...@apple.com> wrote:
> >
> > On 17/12/18 - 08:04:08, Eric Dumazet wrote:
> > > On Fri, Dec 14, 2018 at 2:40 PM Christoph Paasch <cpaa...@apple.com>
> > > wrote:
> > > >
> > >
> > > ...
> > >
> > > > int tcp_fastopen_reset_cipher(struct net *net, struct sock *sk,
> > > > void *key, unsigned int len)
> > > > {
> > > > @@ -96,13 +131,22 @@ error: kfree(ctx);
> > > > spin_lock(&net->ipv4.tcp_fastopen_ctx_lock);
> > > > if (sk) {
> > > > q = &inet_csk(sk)->icsk_accept_queue.fastopenq;
> > >
> > > > + rcu_assign_pointer(ctx->next, q->ctx);
> > > At this point, ctx is not yet visible, so you do not need a barrier yet
> > > ctx->next = q->ctx;
> >
> > Thanks, I will change that.
> >
> > >
> > >
> > > > + rcu_assign_pointer(q->ctx, ctx);
> > >
> > > Note that readers could see 3 contexts in the chain, instead of maximum
> > > two.
> > >
> > > This means that proc_tcp_fastopen_key() (your 3/5 change) would
> > > overflow an automatic array :
> > >
> > > while (ctxt) {
> > > memcpy(&key[i], ctxt->key, TCP_FASTOPEN_KEY_LENGTH);
> > > i += 4;
> > > ctxt = rcu_dereference(ctxt->next);
> > > }
> >
> > Ouch! Thanks for spotting this.
> >
> > If it's ok to have a brief moment of 3 contexts for the readers, I would
> > protect against overflows the readers.
>
> I believe you can refactor the code here, to publish the new pointer
> (rcu_assign_pointer(ctx->next, q->ctx);)
> only after you have shorten the chain.
>
> No worries if one incoming packet can see only the old primary key,
> but not the fallback one,
> since we are anyway about to remove the old fallback.
>
> Ideally the rcu_assign_pointer(ctx->next, q->ctx) operation should be
> the last one, when the new chain
> is clean and ready to be used.
Sounds good, I will do that.
Thanks,
Christoph
