Wireshark-dev: Re: [Wireshark-dev] behavior of tcp_dissect_pdus when protocol pdu is across tcp
From: Sudarshan Raghavan <[email protected]>
Date: Tue, 25 Aug 2009 00:23:33 +0530
I have checked this manually (followed the bytes in the raw TCP stream
in wireshark). I can very clearly see that the start of the new TCP
segment is where the get_pdu_len function pointer is called by
tcp_dissect_pdu. I also verified that the RTMP chunk ends at the
correct offset within the new TCP segment. I have tried with tcp
checksum validation disabled and it still does not work.


On Tue, Aug 25, 2009 at 12:17 AM, wsgd<[email protected]> wrote:
> I think that
> the start of tvb (given to get_pdu_len) corresponds to the start of the pdu.
> So offset will always be zero.
> Olivier
> Sudarshan Raghavan a écrit :
>> I am currently working on adding a RTMP (Real Time Messaging Protocol)
>> dissector to wireshark. RTMP sends out messages by splitting them into
>> chunks. It starts with a default chunk size and sets it to a different
>> value later if required. Each RTMP chunk will contain a chunk header
>> and optionally a message header also.
>> It is possible that a RTMP chunk starts at an offset inside the
>> current TCP segment and spills over to the next TCP segment or later.
>> My length function (get_pdu_len arg of tcp_dissect_pdus) returns the
>> correct value to be able to get hold of the entire chunk. What i am
>> seeing in this case (chunk across TCP segments) is that my length
>> function is getting called as soon as the next TCP segment is seen and
>> the offset argument passed is 0. I was expecting that tcp_dissect_pdus
>> will call the length function at the appropriate offset in the next
>> segment based on the length returned previously. Looking at the
>> implementation of tcp_dissect_pdus in packet-tcp.c seems to confirm my
>> analysis. Am I missing something here? How do I make tcp_dissect_pdus
>> work correctly with chunks across TCP segments.
>> Please note that it works fine if the chunk is contained entirely
>> within a TCP segment.
>> Regards,
>> Sudarshan
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