On Sat, Aug 20, 2016 at 9:47 AM, João Valverde <joao.valverde@tecnico. ulisboa.pt> wrote:
> > On 08/20/2016 02:06 PM, João Valverde wrote: > >> >> On 08/20/2016 02:03 PM, João Valverde wrote: >> >>> I think there is a disconnect here because you are seeing Lua as a >>> system dependency. I see it as Wireshark's own embedded language >>> interpreter (although developed by the Lua team under a suitable >>> license). >>> >>> (Sorry for my brevity, I really appreciate your input). >>> >> >> I also meant to add that it would be better to rip-off the compatibility >> band-aid, IMO, if the 5.3 features justify the upgrade. >> > > I'm not aware of any policy that says our Lua dialect and APIs must be > stable forever. That's nice, I don't like fixing working code as much as > the next person, but sometimes reality intervenes. It would be difficult to > provide that guarantee, seeing as Lua 5.2 is already unmaintained upstreams. I think you're correct that there isn't any policy. Maybe there should be. I really don't have a problem with C API changes. For many years I maintained a set of C dissectors for proprietary protocols that were not really of general interest (and anyway I really did not want to even think about jumping through the hoops that would be necessary to get my employer to allow submitting the changes). I figured that my penance for that "sin" was having to update them regularly as the API changed. (IOW in general I think C dissectors should be contributed to Wireshark--if you don't then you get to live with the pain. :-)) But I think Lua dissectors are different. (I would think) it's a different type of person writing them and the distribution model is different--I think it would be painful if Lua dissector writers have to start saying "this version is compatible with Wireshark 2.0-2.2, this other version is compatible with 2.4-2.6." Would it even make sense to have a Lua equivalent to the VERSION_MAJOR, etc., macros (that I used in C code to make one dissector compatible with several versions of Wireshark)?
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