Question

I created a simple decorator that received an argument (using functions instead of a class), when something strage happened: adding a line a code breaks the execution of a previous line.

Here’s the code:

def my_decorator(sublabel):
    def duration_wrapper(f):

        print sublabel
        # Uncommenting this code will break the previous line - why?
        # if sublabel is None:
        #     sublabel = f.func_name

        def wrapped_function(*args, **kwargs):
            return f(*args, **kwargs)

        return wrapped_function

    return duration_wrapper

@my_decorator('me')
def myf(): pass

myf()

Uncommenting those lines of code cause this exception:

Traceback (most recent call last):
  File "test.py", line 16, in <module>
    @my_decorator('me')
  File "test.py", line 4, in duration_wrapper
    print sublabel
UnboundLocalError: local variable 'sublabel' referenced before assignment

Can anyone explain why uncommenting those 2 lines of code break it?

Answer 1

Decorators are closures, and all labels from an enclosing scope referenced from an enclosed scope must remain static within the enclosed scope. If you want to have a variable from an enclosing scope be mutable from an enclosed scope, you need to wrap it in a mutable object, for example:

def my_decorator(sublabel):
    sublabel = [sublabel]
    def duration_wrapper(f):

        print sublabel[0]
        if sublabel[0] is None:
            sublabel[0] = f.func_name

        def wrapped_function(*args, **kwargs):
            return f(*args, **kwargs)

        return wrapped_function

    return duration_wrapper

As @Bakuriu notes in the comments, Python 3 introduces nonlocal to remove this restriction, and you could also make sublabel a global to get around the problem, but globals are usually a bad idea generally.