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Christopher Rodriguez
Python Development Techdegree Graduate 11,265 PointsReceived Error: UnboundLocalError: local variable 'shopping_list' referenced before assignment
while creating this function I received the error in the subject line.
def add_to_list(item):
shopping_list = shopping_list.append(item)
print("{} has been added to the list! There are {} items in the list.".format(item, len(shopping_list)))
I see where I went wrong. I should have stated:
shopping_list.append(item)
But why is the line shown below incorrect? and what is the error referring to?
shopping_list = shopping_list.append(item)
thanks
2 Answers
Chris Freeman
Treehouse Moderator 68,453 PointsHey Christopher Rodriguez, good question!
The object shopping_list is defined outside the function and is seen as a global from within the function. This changes as soon as an assignment to shopping_list is made within the function. When this is seen by the parser, shopping_list is added to the local namespace and the global version is no longer seen.
Now when shopping_list.append() is run, it is appending to the local shopping_list that has yet to be defined.
This is the same error as went there is a, say, global count variable and count += 1 is used inside a function. count becomes local and hence the right side of count + 1 has an undefined "count".
Post back if you have more questions. Good luck!!
Christopher Rodriguez
Python Development Techdegree Graduate 11,265 PointsI think I understand. so what I am saying in the following line...
shopping_list = shopping_list.append(item)
is...
(Local) shopping_list = (local) shopping_list.append(item)
As opposed to ...
(global) shopping_list.append(item)
Is that correct? Thank you!
Chris Freeman
Treehouse Moderator 68,453 PointsCorrect. Once shopping_list is deemed “local”, all references to it become local.
The kicker is the list.append() method returns None so after the append, None is would be assigned to shopping_list and losing the list.
>>> lst = [1, 2, 3]
>>> lst = lst.append(4)
>>> lst == None
True