[Distutils] distutils.util.get_platform() - Linux vs Windows

samuel.ferencik at barclays.com samuel.ferencik at barclays.com
Tue Aug 20 08:15:46 CEST 2013

>-----Original Message-----
>From: Chris Barker - NOAA Federal [mailto:chris.barker at noaa.gov] 
>Sent: Monday, August 19, 2013 7:13 PM
>To: Ferencik, Samuel: Markets (PRG)
>Cc: distutils-sig at python.org
>Subject: Re: [Distutils] distutils.util.get_platform() - Linux vs Windows
>On Fri, Aug 16, 2013 at 2:18 AM,  <samuel.ferencik at barclays.com> wrote:
>> It seems distutils.util.get_platform() semantically differs on Windows and
>> Linux.
>> Windows: the return value is derived from the architecture of the
>> *interpreter*, hence for 32-bit Python running on 64-bit Windows
>> get_platform() = 'win32' (32-bit).
>> Linux: the return value is derived from the architecture of the *OS*, hence
>> for 32-bit Python running on 64-bit Linux get_platform() = 'linux-x86_64'
>> (64-bit).
>> Is this intentional?
>This seems just plain wrong to me.
>For the record, running a 32 bit Python on a 64 bit OS_X box:
>In [5]: distutils.util.get_platform()
>Out[5]: 'macosx-10.6-i386'
>which is the answer I want.


What does your 'uname -m' return? Is it possible you're really running a 32-bit
Python on a *32-bit* OS X kernel? [http://superuser.com/q/161195]

Basically, you're saying that the return value is wrong on Linux and correct on
Windows, right? That get_platform() should return "32-bit" for a 32-bit process
running on a 64-bit system. TBH, I was expecting the opposite; to me, "platform"
means the OS, which would mean that Linux does well to derive the return value
from the OS's architecture.



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