[Tutor] why does platform.architecture default to sys.executable?

[Albert-Jan Roskam]

-------------------------------------------
On Sat, 10/26/13, Amit Saha <amitsaha.in at gmail.com> wrote:

 Subject: Re: [Tutor] why does platform.architecture default to sys.executable?
 To: "Albert-Jan Roskam" <fomcl at yahoo.com>
 Cc: "Python Mailing List" <tutor at python.org>
 Date: Saturday, October 26, 2013, 6:51 PM
 
 On Sun, Oct 27, 2013 at 2:39 AM,
 Albert-Jan Roskam <fomcl at yahoo.com>
 wrote:
 > Hi,
 >
 > Why does the "executable" parameter default to
 sys.executable? Yesterday I
 > was surprised to see platform.architecture return
 "32bit" on a 64-bit
 > system, just because a 32-bit Python interpreter was
 installed. Wouldn't
 > this make more sense:
 >
 > import sys, platform
 > pf = sys.platform.lower()[:3]
 > executable = "iexplore.exe" if pf[:3] == "win" else
 "/bin/ls"
 
 I think it's mainly because of avoiding choosing arbitrary
 programs,
 although they are most certainly guaranteed to be present.
 Besides,
 there are better ways to find the platform architecture, I
 think.
 os.uname() comes to mind.
 

===>  os.uname is Unix-only:

http://docs.python.org/2/library/os.html#os.uname
os.uname()

    Return a 5-tuple containing information identifying the current operating system. The tuple contains 5 strings: (sysname, nodename, release, version, machine). Some systems truncate the nodename to 8 characters or to the leading component; a better way to get the hostname is socket.gethostname() or even socket.gethostbyaddr(socket.gethostname()).

    Availability: recent flavors of Unix.