Wireshark-dev: Re: [Wireshark-dev] Distributing a wireshark dissector
From: "Anders Broman" <anders.broman@xxxxxxxxxxxx>
Date: Thu, 15 Nov 2007 16:16:51 +0100
Hi,
Do you build with VC6? If not that's probably the cause of the problem.
 
Regards
Anders


From: wireshark-dev-bounces@xxxxxxxxxxxxx [mailto:wireshark-dev-bounces@xxxxxxxxxxxxx] On Behalf Of Eytan Kidron
Sent: den 15 november 2007 15:21
To: wireshark-dev@xxxxxxxxxxxxx
Subject: [Wireshark-dev] Distributing a wireshark dissector

 

Hi,

 

I have a question concerning the creation and distribution of dissectors.

I created a dissector using a tutorial which I found in http://www.codeproject.com/useritems/custom_dissector.asp, and now I have a dll called my-protocol.dll.

I am not interested, at this point, in distributing this dissector to the general public, but I am interested in distributing it within my organization. I found that I could do that by creating a wireshark installer (see step 12 in the link above).

When others in my organization executed this installer and copied the dll to the appropriate place, they could use the dissector. The problem with that is that whenever a new version of wireshark is released, my dissector will obviously not be there, and if somebody wants some feature from a newer wireshark version and also wants my dissector, they need two separate wireshark installations.

 

My question is:

Is it possible to add a new dissector to an existing wireshark without having to reinstall it or install a separate instance of wireshark?

 

One might expect that I could simply copy my-protocol.dll to the folder where all the other dlls are, but when I tried to do this I got an error in loading a C runtime library (incidentally, I am working on Windows XP). Is that supposed to work?

 

I don't mind the users in my organization having to do something extra (like copying a dll) every time they update their wireshark, if they want to use this dissector, but I don’t want to need to create a new installer whenever a new wireshark version is released.

 

I would be grateful if anyone could help me with this.

Thanks,

Eytan