function - Unbound local variable problem in Python -

i've got following code snippet:

def isolation_level(level):     def decorator(fn):         def recur(level, *args, **kwargs):             if connection.inside_block:                 if connection.isolation_level < level:                     raise isolationlevelerror(connection)                 else:                     fn(*args, **kwargs)             else:                 connection.enter_block()                 try:                     connection.set_isolation_level(level)                     fn(*args, **kwargs)                     connection.commit()                 except isolationlevelerror, e:                     connection.rollback()                     recur(e.level, *args, **kwargs)                 finally:                     connection.leave_block()         def newfn(*args, **kwargs):             if level none: # <<<< error message here, unbound local variable `level`                 if len(args):                     if hasattr(args[0], 'isolation_level'):                         level = args[0].isolation_level                 elif kwargs.has_key('self'):                     if hasattr(kwargs['self'], 'isolation_level'):                         level = kwargs.pop('self', 1)              if connection.is_dirty():                 connection.commit()             recur(level, *args, **kwargs)         return newfn     return decorator 

it doesn't matter does, post in original form, unable recreate situation simpler.

the problem when call isolation_level(1)(some_func)(some, args, here) unbound local variable exception in line 21 (marked on listing). don't understand why. tried recreating same structure of functions , function calls wouldn't contain implementation details figure out wrong. don't exception message then. example following works:

def outer(x=none):     def outer2(y):         def inner(x, *args, **kwargs):             print x             print y             print args             print kwargs         def inner2(*args, **kwargs):             if x none:                 print "i'm confused"             inner(x, *args, **kwargs)         return inner2     return outer2  outer(1)(2)(3, z=4) 

prints:

1 2 (3,) {'z': 4} 

what missing??

edit

ok, problem in first version perform assignment variable. python detects , therefore assumes variable local.

local variables determined @ compile time: assignments variable level few lines below line error occurs in make variable local inner function. line

if level none: 

actually tries access variable level in innermost scope, such variable not exist yet. in python 3.x, can solve problem declaring

nonlocal level 

at beginning of inner function, if want alter variable of outer function. otherwise, can use different variable name in inner function.