netdev - Re: [PATCH net-next] tcp: avoid RST in 3-way shakehands due to failure in tcp_timewait_state

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Message-ID: <CAL+tcoBA78svT_vTMOLV-pbwKM1o_SDbjs7AAZLhHOtrd8akBg@mail.gmail.com>
Date: Thu, 7 Nov 2024 16:26:31 +0800
From: Jason Xing <kerneljasonxing@...il.com>
To: Philo Lu <lulie@...ux.alibaba.com>
Cc: davem@...emloft.net, edumazet@...gle.com, kuba@...nel.org, 
	pabeni@...hat.com, dsahern@...nel.org, horms@...nel.org, 
	netdev@...r.kernel.org, Jason Xing <kernelxing@...cent.com>
Subject: Re: [PATCH net-next] tcp: avoid RST in 3-way shakehands due to
 failure in tcp_timewait_state_process

On Thu, Nov 7, 2024 at 4:22 PM Philo Lu <lulie@...ux.alibaba.com> wrote:
>
>
>
> On 2024/11/7 16:01, Jason Xing wrote:
> > On Thu, Nov 7, 2024 at 3:51 PM Philo Lu <lulie@...ux.alibaba.com> wrote:
> >>
> >> Hi Jason,
> >>
> >> On 2024/11/5 10:55, Jason Xing wrote:
> >>> From: Jason Xing <kernelxing@...cent.com>
> >>>
> >>> We found there are rare chances that some RST packets appear during
> >>> the shakehands because the timewait socket cannot accept the SYN and
> >>> doesn't return TCP_TW_SYN in tcp_timewait_state_process().
> >>>
> >>> Here is how things happen in production:
> >>> Time        Client(A)        Server(B)
> >>> 0s          SYN-->
> >>> ...
> >>> 132s                         <-- FIN
> >>> ...
> >>> 169s        FIN-->
> >>> 169s                         <-- ACK
> >>> 169s        SYN-->
> >>> 169s                         <-- ACK
> >>> 169s        RST-->
> >>> As above picture shows, the two flows have a start time difference
> >>> of 169 seconds. B starts to send FIN so it will finally enter into
> >>> TIMEWAIT state. Nearly at the same time A launches a new connection
> >>> that soon is reset by itself due to receiving a ACK.
> >>>
> >>> There are two key checks in tcp_timewait_state_process() when timewait
> >>> socket in B receives the SYN packet:
> >>> 1) after(TCP_SKB_CB(skb)->seq, rcv_nxt)
> >>> 2) (s32)(READ_ONCE(tcptw->tw_ts_recent) - tmp_opt.rcv_tsval) < 0)
> >>>
> >>> Regarding the first rule, it fails as expected because in the first
> >>> connection the seq of SYN sent from A is 1892994276, then 169s have
> >>> passed, the second SYN has 239034613 (caused by overflow of s32).
> >>>
> >>> Then how about the second rule?
> >>> It fails again!
> >>> Let's take a look at how the tsval comes out:
> >>> __tcp_transmit_skb()
> >>>       -> tcp_syn_options()
> >>>           -> opts->tsval = tcp_skb_timestamp_ts(tp->tcp_usec_ts, skb) + tp->tsoffset;
> >>> The timestamp depends on two things, one is skb->skb_mstamp_ns, the
> >>> other is tp->tsoffset. The latter value is fixed, so we don't need
> >>> to care about it. If both operations (sending FIN and then starting
> >>> sending SYN) from A happen in 1ms, then the tsval would be the same.
> >>> It can be clearly seen in the tcpdump log. Notice that the tsval is
> >>> with millisecond precision.
> >>>
> >>> Based on the above analysis, I decided to make a small change to
> >>> the check in tcp_timewait_state_process() so that the second flow
> >>> would not fail.
> >>>
> >>
> >> I wonder what a bad result the RST causes. As far as I know, the client
> >> will not close the connect and return. Instead, it re-sends an SYN in
> >> TCP_TIMEOUT_MIN(2) jiffies (implemented in
> >> tcp_rcv_synsent_state_process). So the second connection could still be
> >> established successfully, at the cost of a bit more delay. Like:
> >>
> >>    Time        Client(A)        Server(B)
> >>    0s          SYN-->
> >>    ...
> >>    132s                         <-- FIN
> >>    ...
> >>    169s        FIN-->
> >>    169s                         <-- ACK
> >>    169s        SYN-->
> >>    169s                         <-- ACK
> >>    169s        RST-->
> >> ~2jiffies    SYN-->
> >>                                 <-- SYN,ACK
> >
> > That's exactly what I meant here :) Originally I didn't expect the
> > application to relaunch a connection in this case.
>
> s/application/kernel/, right?

No. Perhaps I didn't make myself clear. If the kernel doesn't silently
drop the SYN and then send back an ACK, the application has to call
the connect() syscall again.

> Because the retry is transparent to user
> applications except the additional latency. I think all of these are
> finished in a single connect() :)

Right.