Major Dávid <major.david@xxxxxxxxxx> wrote: > On 3/3/23 01:09, Florian Westphal wrote: > > > > Which one? As far as I can see TCP stack would end up adding a > > duplicate quadruple to the hash if we only drop the reference and > > keep the listen sk around. > > I just thought that tcp_timewait_state_process is called by TCP stack to > handle TW state, which actually call inet_twsk_deschedule_put parallel to tproxy and > that would be the root cause of the deadlock. No, it won't be called. We can do two things: 1. Assign the tw sk to skb->sk, then its handled by tcp_timewait_state_process() in tcp stack. Problem is that after the tw sk was handled, tcp stack won't find a listener socket if the tproxy service is running on a different port. 2. Assign the listener socket to skb->sk (this is whats done now). > So I guess now, basically we would leak away the tw socket if we do not call put in tproxy? We could just drop the reference, but then, as far as i can see, we end up with two identical connection entries in the ehash table.
