Scope of variables in python decorator

Question

I'm having a very weird problem in a Python 3 decorator.

If I do this:

def rounds(nr_of_rounds):
    def wrapper(func):
        @wraps(func)
        def inner(*args, **kwargs):
            return nr_of_rounds
        return inner
    return wrapper

it works just fine. However, if I do this:

def rounds(nr_of_rounds):
    def wrapper(func):
        @wraps(func)
        def inner(*args, **kwargs):
            lst = []
            while nr_of_rounds > 0:
                lst.append(func(*args, **kwargs))
                nr_of_rounds -= 1
            return max(lst)
        return inner
    return wrapper

I get:

while nr_of_rounds > 0:
UnboundLocalError: local variable 'nr_of_rounds' referenced before assignment

In other words, I can use nr_of_roundsin the inner function if I use it in a return, but I can't do anything else with it. Why is that?

This seems possibly hazardous: the first call to the wrapped function will decrement the number of rounds to zero, after which the round count will always be zero. Perhaps you want to initialize a local counter instead? — nneonneo
– nneonneo, Commented Apr 21, 2015 at 0:25

mgilson · Accepted Answer · 2015-04-21 14:54:18Z

16

Since nr_of_rounds is picked up by the closure, you can think of it as a "read-only" variable. If you want to write to it (e.g. to decrement it), you need to tell python explicitly -- In this case, the python3.x nonlocal keyword would work.

As a brief explanation, what Cpython does when it encounters a function definition is it looks at the code and decides if all the variables are local or non-local. Local variables (by default) are anything that appear on the left-hand side of an assignment statement, loop variables and the input arguments. Every other name is non-local. This allows some neat optimizations¹. To use a non-local variable the same way you would a local, you need to tell python explicitly either via a global or nonlocal statement. When python encounters something that it thinks should be a local, but really isn't, you get an UnboundLocalError.

^{¹The Cpython bytecode generator turns the local names into indices in an array so that local name lookup (the LOAD_FAST bytecode instruction) is as fast as indexing an array plus the normal bytecode overhead.}

edited Apr 21, 2015 at 14:54

answered Apr 21, 2015 at 0:20

mgilson

312k70 gold badges656 silver badges722 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Ben Over a year ago

Thanks for your answer; it did solve the issue. What is the difference between a global and nonlocal? global does not seem to work in this case.

mgilson Over a year ago

Nope, a global must live in the global namespace (e.g. at the module level). nonlocal means it could be either global or picked up via a closure.

Veedrac Over a year ago

nonlocal won't pick up globals.

mgilson Over a year ago

@Veedrac -- You're correct, non-local won't pick up globals. My comment was incorrect.

Ashwini Chaudhary Over a year ago

We can call nr_of_rounds a free variable.

Community · Accepted Answer · 2017-05-23 12:07:12Z

0

Currently there is no way to do the same for variables in enclosing function scopes, but Python 3 introduces a new keyword, "nonlocal" which will act in a similar way to global, but for nested function scopes.
so in your case just use like:
def inner(*args, **kwargs): nonlocal nr_of_rounds lst = [] while nr_of_rounds > 0: lst.append(func(*args, **kwargs)) nr_of_rounds -= 1 return max(lst) return inner
For more info Short Description of the Scoping Rules?

edited May 23, 2017 at 12:07

CommunityBot

11 silver badge

answered Apr 25, 2015 at 19:11

GrvTyagi

4,5071 gold badge37 silver badges42 bronze badges

Collectives™ on Stack Overflow

Scope of variables in python decorator

2 Answers 2

5 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

5 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related