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From: Henry Olders <henry.olders@mcgill.ca>
Subject: scope of function parameters
Date: Sun, 29 May 2011 04:30:52 -0400
To: python-list@python.org
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Newsgroups: comp.lang.python
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:6521

I just spent a considerable amount of time and effort debugging a program. The made-up code snippet below illustrates the problem I encountered:

def main():
	a = ['a list','with','three elements']
	print a
	print fnc1(a)
	print a
	
def fnc1(b):
	return fnc2(b)

def fnc2(c):
	c[1] = 'having'
	return c

This is the output:
['a list', 'with', 'three elements']
['a list', 'having', 'three elements']
['a list', 'having', 'three elements']

I had expected the third print statement to give the same output as the first, but variable a had been changed by changing variable c in fnc2.

It seems that in Python, a variable inside a function is global unless it's assigned. This rule has apparently been adopted in order to reduce clutter by not having to have global declarations all over the place.

I would have thought that a function parameter would automatically be considered local to the function. It doesn't make sense to me to pass a global to a function as a parameter.

One workaround is to call a function with a copy of the list, eg in fnc1 I would have the statement "return fnc2(b[:]". But this seems ugly.

Are there others who feel as I do that a function parameter should always be local to the function? Or am I missing something here?

Henry