[Zope-Coders] binary releases
Matthew T. Kromer
matt@zope.com
Fri, 2 Nov 2001 08:27:32 -0500
On Thursday, November 1, 2001, at 05:20 PM, Martijn Faassen wrote:
> Hi there,
>
> I'd like to release a new development version of ParsedXML. One
> complication is that ParsedXML comes with an Expat binary. For the
> source version this is not a problem, but Windows users are used to
> binary versions. Since we're also still supporting 1.5.2 I need two
> releases for Windows, one for Zope 2.4/Python 2.1 and one for Zope 2.3/
> Python 1.5.2 (note that the current CVS was mildly broken for Python
> 1.5.2
> due to a missing #include <assert.h> in pyexpat.c, in case you feel like
> surprising me with a binary :)
>
> Who creates these windows binaries? I don't have a Windows compiler,
> myself.
> What kind of release procedures do we have in the first place? Where
> would
> I put the release? How does a release become 'official', and how
> official
> should it be?
Hi Martijn,
Usual either Brian Lloyd or I create the windows binary releases. It's
actually fairly straightforward; it just requires Visual C++ 6.0 to
build; there's a script that does most of the heavy lifting. After we
make the binary objects, they get put into a special internal CVS tree
keeping the binaries; the release builder script uses that to crank out
a release.
A final step is taking the output of the release builder and running it
through the WISE packaging engine to create the installer.
However, I don't think we'd contemplated the notion of having another
2.3 release (for Python 1.5.2) since 2.4 is the current stable release
branch.
Having said that, what are YOUR requirements, exactly w.r.t doing a new
release? Do you want a 2.3.X public release, or do you just want to get
your hands on a windows installer file which contains your fixes? I can
make arguments either way, e.g. if we go through the hassle of building
another release, everyone should be able to share BUT we want to put
people on the Python 2.1 path and not continue to develop on the 1.5.2
path.