Modifying a variable in a non-global outer scope?

[Edward C. Jones]

#! /usr/bin/env python
"""
When I run the following program I get the error message:

UnboundLocalErr or: local variable 'x' referenced before assignment

Can "inner" change the value of a variable defined in "outer"? Where
is this explained in the docs?
"""
def outer():
def inner():
x = x + 1

x = 3
inner()
print x

outer()

[bruno at modulix]

Re: Modifying a variable in a non-global outer scope?

Edward C. Jones wrote:[color=blue]
> #! /usr/bin/env python
> """
> When I run the following program I get the error message:
>
> UnboundLocalErr or: local variable 'x' referenced before assignment
>
> Can "inner" change the value of a variable defined in "outer"?[/color]

Not this way
[color=blue]
> Where
> is this explained in the docs?[/color]

IIRC,

http://www.python.org/doc/2.4.2/ref/naming.html

[color=blue]
> """
> def outer():
> def inner():
> x = x + 1
>
> x = 3
> inner()
> print x
>
> outer()[/color]

What are functions arguments and return values for ?

def outer():
def inner(x):
return x+1
x = 3
x = inner(x)
print x

outer()

Using side-effects - specially this way - is a Very Bad Thing(tm). It
makes code that is hard to read and hard to maintain.


--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in 'onurb@xiludom. gro'.split('@')])"