Lexical Scope

From: Matt Knepley (knepley_at_mcs.anl.gov)
Date: 10/30/03 

Date: 30 Oct 2003 07:59:04 -0800

I must be misunderstanding how Python 2.3 handles lexical scoping.
Here is a sample piece of code:

def run():
  a = 1
  def run2(b):
    print a
  run2(2)
  print a
run()

which gives the output:

1
1

whereas this piece of code:

def run():
  a = 1
  def run2(b):
    a = b
    print a
  run2(2)
  print a
run()

gives:

2
1

and finally this code bombs:

def run():
  a = 1
  def run2(b):
    print a
    a = b
  run2(2)
  print a
run()

with an error about UnboundLocal. It seems that lexical scope works
only for references, and as soon as I make an assignment a new local
is created. Is this true?

    Matt

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