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:

UnboundLocalError: 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]

Edward C. Jones wrote:

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

UnboundLocalError: local variable 'x' referenced before assignment

Can "inner" change the value of a variable defined in "outer"?
Not this way
Where
is this explained in the docs?
IIRC,
http://www.python.org/doc/2.4.2/ref/naming.html
"""
def outer():
def inner():
x = x + 1

x = 3
inner()
print x

outer()

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 'o****@xiludom.gro'.split('@')])"