function parameter scope python 2.5.2

[George Sakkis]

On Nov 20, 6:40 pm, J Kenneth King <ja... at agentultra.com> wrote:
> J Kenneth King <ja... at agentultra.com> writes:
>
>
>
> > I recently encountered some interesting behaviour that looks like a bug
> > to me, but I can't find the appropriate reference to any specifications
> > to clarify whether it is a bug.
>
> > Here's the example code to demonstrate the issue:
>
> > class SomeObject(object):
>
> >     def __init__(self):
> >         self.words = ['one', 'two', 'three', 'four', 'five']
>
> >     def main(self):
> >         recursive_func(self.words)
> >         print self.words
>
> > def recursive_func(words):
> >     if len(words) > 0:
> >         word = words.pop()
> >         print "Popped: %s" % word
> >         recursive_func(words)
> >     else:
> >         print "Done"
>
> > if __name__ == '__main__':
> >     weird_obj = SomeObject()
> >     weird_obj.main()
>
> > The output is:
>
> > Popped: five
> > Popped: four
> > Popped: three
> > Popped: two
> > Popped: one
> > Done
> > []
>
> > Of course I expected that recursive_func() would receive a copy of
> > weird_obj.words but it appears to happily modify the object.
>
> > Of course a work around is to explicitly create a copy of the object
> > property befor passing it to recursive_func, but if it's used more than
> > once inside various parts of the class that could get messy.
>
> > Any thoughts? Am I crazy and this is supposed to be the way python works?
>
> Of course, providing a shallow (or deep as necessary) copy makes it
> work, I'm curious as to why the value passed as a parameter to a
> function outside the class is passed a reference rather than a copy.

Why should it be a copy by default ? In Python all copies have to be
explicit.

George