Re: [lwip-users] TCP retransmission

[address@hidden]

Nikolas Karakotas wrote:
> True for not providing enough information.
>
> Deducing wireshark information I can say:
> 1. Remote host send a Modbus request
> 2. For whichever reason we don't respond to it as it may have been
> lost because of network or Ethernet driver
> 3. The remote host then re sends the request and we are able to
> receive it notifying with a Dup Ack
> 4. The remote host then sends the re-transmission of the previous
> packets lost with 24 bytes ( contains 2 modbus requests now)
> 5. It's forwarded to the application layer but I found out that the
> application layer Modbus implementation doesn't handle fragmented
> Modbus requests well

Please, more info! I can only assume that lwIP is the Modbus-TCP server 
having the IP address 10.200.61.1 and 10.200.221.100 is some non-lwIP 
Modbus-TCP client.
In that case, it seems like the Modbus-TCP client retransmits 2 requests 
in one TCP segment (note that frame 2 and 3 have a TCP length of 12 
while frame 5 has 24).
This is perfectly valid for TCP. It seems like your application cannot 
handle 2 requests coming in with one "recv" callback.

This is a common error people do when programming TCP applications: you 
get a stream, not packets. You need to handle the depacketization at 
application level.

Simon