Re: pyROOT on Mac OS X Snow Leopard

Hi,

I have 10.6.1, and he say:

higgs-phy-bris-ac-uk:~ phdmn$ file `which python`

/usr/bin/python: Mach-O universal binary with 3 architectures
/usr/bin/python (for architecture x86_64):	Mach-O 64-bit executable x86_64
/usr/bin/python (for architecture i386):	Mach-O executable i386
/usr/bin/python (for architecture ppc7400):	Mach-O executable ppc

   Dave

James Jackson wrote:
> Hi,
>
> ROOT will have been built for 64 bit architecture, I bet. On my machine,
> I have the following:
>
> Unknown-00-21-e9-e2-86-6e:~ jamesjackson$ file `which root`
> /usr/local/bin/root: Mach-O 64-bit executable x86_64
>
> Unknown-00-21-e9-e2-86-6e:~ jamesjackson$ file `which python`
> /usr/bin/python: Mach-O universal binary with 2 architectures
> /usr/bin/python (for architecture ppc7400): Mach-O executable ppc
> /usr/bin/python (for architecture i386): Mach-O executable i386
>
> One could possibly force ROOT to build for 32 bit, or re-build python
> for 64 bit. I suspect the former is easier...
>
> Regards,
> James.
>
> On 30 Oct 2009, at 08:33, federico roncarolo wrote:
>

>> Hello,
>>
>> I just installed Mac OS X 10.6 (Snow Leopard), python 2.6, gcc 4.2.1, 
>> and compiled ROOT 5.25/02 with architecture = macosx.
>>
>> Exactely as in the post:
>>
>> <http://root.cern.ch/phpBB2/viewtopic.php?p=39830&sid=7516d970f76e5ebde824f5a0a11dca03> 
>>
>>
>> when importing ROOT from python I get:
>>
>> [imac]:-) > python
>> Python 2.6.1 (r261:67515, Jul  7 2009, 23:51:51)
>> [GCC 4.2.1 (Apple Inc. build 5646)] on darwin
>> Type "help", "copyright", "credits" or "license" for more information.
>> >>> import ROOT
>> Traceback (most recent call last):
>> File "<stdin>", line 1, in <module>
>> File "lib/ROOT.py", line 86, in <module>
>>   import libPyROOT as _root
>> ImportError: dlopen(/root/lib/root/libPyROOT.so, 2): no suitable image 
>> found.  Did find:
>>     /root/lib/root/libPyROOT.so: mach-o, but wrong architecture
>>
>>
>> Is there a solution?
>> Should I compile another version of ROOT or of Python?
>>
>> Many thanks,
>> Federico
>>