I have the 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 is None:
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 does not matter what he does, however I post it in the original form, as I was not able to recreate the situation with anything simpler.
The problem is that when I call isolation_level(1)(some_func)(some, args, here), I get an exception Unbound local variableon line 21 (marked in the listing). I do not understand why. I tried to recreate the same structure of functions and function calls that do not contain all implementation details in order to find out what is wrong. However, I do not receive an exception message. For example, the 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 is None:
print "I'm confused"
inner(x, *args, **kwargs)
return inner2
return outer2
outer(1)(2)(3, z=4)
Print
1
2
(3,)
{'z': 4}
What am I missing?
EDIT
So, the problem is that in the first version I actually perform the assignment of a variable. Python detects this and therefore assumes the variable is local.