[Pythonmac-SIG] [Numpy-discussion] Mac OSX 4-way universal Re: python 2.6 trunk

Chris Kees christopher.e.kees at usace.army.mil
Thu Aug 14 20:38:54 CEST 2008


The 4-way universal install of numpy-1.1.1 is working now with the Python
2.6b2+ (trunk:65678), and  all the tests pass (running as i386 and x86_64 at
least). Unfortunately, I didn't find exactly what  was  causing it. I just
erased /Library/Frameworks/Python64.framework and rebuilt the 4-way
universal python and numpy module again because I had noticed that  the
numpy build directory kept coming up with this structure:

%ls build
lib.macosx-10.5-universal-2.6        scripts.macosx-10.5-universal-2.6
src.macosx-10.3-i386-2.6        temp.macosx-10.5-universal-2.6

Apparently something in my python framework was leftover from a previous
install and causing the src.macosx-10.3-i386-2.6 to get built, which seems
to be related to distutils deciding that I was cross-compiling. Anyway,
thanks  for the help and sorry for the trouble. I guess one should always
erase the python framework before re-installing it? Should I be running an
uninstall script instead  of just erasing it?

Chris


On 8/13/08 6:08 PM, "robert.kern at gmail.com" <robert.kern at gmail.com> wrote:

> On 2008-08-13, David Cournapeau <cournape at gmail.com> wrote:
>> On Wed, Aug 13, 2008 at 4:20 PM, Robert Kern <robert.kern at gmail.com> wrote:
>>> 
>>> Hmm. Odd. I can't find the string "Can't install when cross-compiling"
>>> anywhere in the numpy or Python sources. Can you try again with the
>>> environment variable DISTUTILS_DEBUG=1 set?
>> 
>> You can find it in python svn: the message seems python 2.6 specific.
> 
> Okay, it looks like this happens when distutils.util.get_platform()
> and the build command's plat_name are different. Chris, can you do the
> following and show me the output?
> 
> $ python setup.py build --help
> ...
> $ python -c "from distutils import util;print util.get_platform()"
> ...
> 
> Probably a workaround is to do
> 
> $ python setup.py build --plat-name=... install
> 
> where ... is whatever the output of the second command above gives.



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