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

This is not a bug, just a common gotcha.  The rules are
described at:

http://docs.python.org/reference/executionmodel.html#naming

Here's the relevant excerpt:

"""If a name binding operation occurs anywhere within a code block, all 
uses of the name within the block are treated as references to the 
current block. This can lead to errors when a name is used within a 
block before it is bound. This rule is subtle. Python lacks declarations 
and allows name binding operations to occur anywhere within a code 
block. The local variables of a code block can be determined by scanning 
the entire text of the block for name binding operations."""

Closing as invalid.