New submission from vid podpecan <[email protected]>:
Consider the following two functions:
def outer():
a = 1
def inner():
print a
inner()
#end outer()
def outer_BUG():
a = 1
def inner():
print a
a = 2
inner()
#end outer_BUG()
The first function outer() works as expected (it prints 1), but the
second function ends with an UnboundLocalError, which says that the
"print a" statement inside inner() function references a variable before
assignment. Somehow, the interpreter gets to this conclusion by looking
at the next statement (a = 2) and forgets the already present variable a
from outer function.
This was observed with python 2.5.4 and older 2.5.2. Other releases were
not inspected.
Best regards,
Vid
----------
messages: 85992
nosy: vpodpecan
severity: normal
status: open
title: scope resolving error
type: behavior
versions: Python 2.5
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Python tracker <[email protected]>
<http://bugs.python.org/issue5763>
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