Python scoping: understading LEGB

Update: Thanks to Avazu for a clean suggestion and Fred for pointing out I should have indicated a good way to get around this.


Python scoping fun! Read about LEGB to understand the basics of python scoping.


I never bothered to read about how python scoping works until I hit this. It's not exactly something to research until you have issues with it. :)

I had something like this going on:

def func1(param=None):
    def func2():
        if not param:
            param = 'default'
        print param
    # Just return func2.
    return func2

if __name__ == '__main__':

Note: Actual code was not as straightforward, func2 was actually a decorator. Admittedly, using the same parameter name is not a must, but it's still a curiosity. I just wanted to fall back to a default value on run-time.

If you try to run this in python, here's what you get:

~ $ python 
Traceback (most recent call last):
  File "", line 11, in <module>
  File "", line 3, in func2
    if not param:
UnboundLocalError: local variable 'param' referenced before assignment

If you're curious, you can read about the principles of LEGB. You have to understand a bit about compilers and the AST to get what's going on behind the scenes. You might think that replacing lines 3-4 with:

param = param or 'default'

Might work. But no. You can't assign the same parameter at the local level if the enclosing level defines it. Even this fails:

param = param

Fun, no?

What to do?

There are a few ways to get around this.

  • Assign param outside of func2. This doesn't work if you need the default value to be dependent on what params func2 receives.
  • Use a second variable, param2 inside of func2 (posted below).

Here is the solution suggested by our commenter Avazu (over on the mozilla webdev blog):

def func1(param=None):
    def func2(param2=param):
        if not param2:
            param2 = 'default'
        print param2
    # Just return func2.
    return func2

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