scope of function parameters

[Mel]

Henry Olders wrote:

> 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.

It doesn't look like a question of local or global.  fnc2 is passed a 
container object and replaces item 1 in that container.  You see the results 
when fnc2 prints the object it knows as `c`, and you see again when main 
prints the object it knows as `a`.  Python doesn't pass parameters by 
handing around copies that can be thought of as local or global.  Python 
passes parameters by binding objects to names in the callee's namespace.  In 
your program the list known as `a` in main is identically the same list as 
the one known as `c` in fnc2, and what happens happens.

	Mel.