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

[Albert-Jan Roskam]

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"
arch = platform.architecture(executable)[0]
 

Regards, 
Albert-Jan 

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