[Stackless] Pre 2.5.2 merge / testing help appreciated

[Jeff Senn]

Seems to work on OS-X (10.5.1) -- with a few tiny caveats which I'll  
mention
for the record, but don't seem terribly relevant to anyone not trying  
to build
on OS-X.

There were a few problems building it using the build script - but I  
attribute
that to a non-released version (e.g. one problem was fetching the html  
for doing
a documentation build from the python site, another was fetching a  
particular version
of sqlite).

The 2 test weirdnesses were:

   test_subprocess now triggers the OS-X "the Application Unexpectedly  
Quit" dialog
   box that the system pops up when a process (that is considered an  
"Application") dies for certain reasons.
   This could be completely normal due to how the test is terminating  
the process.

   test_socket failed once.  But I have no more details since it now  
succeeds every time
   I run it and it seemed innocuous, so I'm sticking my head back into  
the sand on that one.

      File "/Library/Frameworks/Python.framework/Versions/2.5/lib/ 
python2.5/test/test_socket.py", line 122, in _tearDown
     self.fail(msg)
AssertionError: (57, 'Socket is not connected')


-Jas



On Jan 31, 2008, at 12:05 PM, Richard Tew wrote:

> Hi,
>
> I have merged in the current state of the mainline release25maint
> branch and we have something which comes up as 2.5.2a0.
>
> I have not regenerated the configure file.  I need to do this on a
> Linux machine.
> I have not tested it with all the dependencies installed (ssl, zlib,
> etc).  Can someone do this?
> I have tested it otherwise.
>
> If people who have the time and interest can get it, compile it and
> test it, it would be appreciated:
> http://svn.python.org/projects/stackless/branches/release25-maint/
>
> Test the regression tests, run Lib/test/regrtest.py
> Test the Stackless unit tests, cd into the Stackless/unittests source
> directory and run runall.py with the python executable you have built
> with this source code.
>
> Please let me know how it goes :-)
>
> Thanks,
> Richard.
>
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