How to get the file path of a module from a function executed but not declared in it?

Question

If I want the path of the current module, I\'ll use __file__.

Now let\'s say I want a function to return that. I can\'t do:

def get_path         


        

Answer 1

This is how I would do it:

import sys

def get_path():
    namespace = sys._getframe(1).f_globals  # caller's globals
    return namespace.get('__file__')