Note: this is a repost from my blog. You can find the original post here.
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__': func1('test')()
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 test.py Traceback (most recent call last): File "test.py", line 11, in
func1('test')() File "test.py", 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
What to do?
There are a few ways to get around this.
paramoutside of func2. This doesn’t work if you need the default value to be dependent on what params func2 receives.
- Use a second variable,
param2inside of func2 (posted below).
Here is the solution suggested by our commenter Avazu:
def func1(param=None): def func2(param2=param): if not param2: param2 = 'default' print param2 # Just return func2. return func2