How Do I fix an UnboundLocalError in Python 3?

Question

todolist = []

def add_item(item):
    todolist =  todolist + [item]

def main():

    add_item(1)

    print(todolist)


if __name__ == '__main__':
    main()

I am trying to make a function called add_item() which works like append() and I am not allowed to use any built in functions. I keep getting an UnboundLocalError. How would I fix this?

Your constraints are weird. All operators including object creation are effectively built in functions, and Python is an imperative language to the degree that even `def` is an executable statement. `global` is a rare exception in that it's a declarative directive for the compiler. — Yann Vernier, Nov 18 '17 at 14:54

score 1 · Answer 1 · answered Nov 18 '17 at 15:21

Even if you fix the local variable issue, your code doesn't behave like list.append. append operates by stateful side effect, mutating the list it was run on; your code created a new list and assigned a name. The only way I can think of to mutate a list that way without using a named method is a slice assignment:

def myappend(intolist, newitem):
    intolist[len(intolist):] = [newitem]

But this obviously uses the len built in function, and the assignment is translated into a setitem call. It's possible to avoid using len, by using implicit bool and getitem calls. But the calls are still there; basically, only a program that performs no operations can run without calling built in functions.

Hey I just made a function call size() that returns the size of the list instead of using len, and thanks very much. — blazing, Nov 18 '17 at 15:39

Abhijith Asokan · Answer 2 · 2017-11-18T13:14:02.073

0

Because the statement todolist = todolist + [item], which is an assignment statement that creates a local variable todolist hides the global variable with the same name. So you have to specify that the variable is in global scope using the keyword global.

def add_item(item):
    global todolist
    todolist =  todolist + [item]

When you use append(),

 todolist.append(item)

there is no assignment operation, hence no variable is created and the variable in the global scope is used.

edited Nov 18 '17 at 13:14

answered Nov 18 '17 at 13:07

Abhijith Asokan

1,865
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Is there any other was of doing it, without using global? The program must be completely imperative. – blazing Nov 18 '17 at 13:30
@blazing use `append()` instead of `+`. – Abhijith Asokan Nov 18 '17 at 13:32
That would make it not completely imperative though, because append calls from the List class. – blazing Nov 18 '17 at 13:35
@blazing other way is, you could pass the list `todolist` as an argument to the function. – Abhijith Asokan Nov 18 '17 at 13:48
The list.append method is [imperative](https://en.wikipedia.org/wiki/Imperative_programming). OO, imperative and procedural are all coexisting [programming paradigms](https://en.wikipedia.org/wiki/Programming_paradigm). – Yann Vernier Nov 18 '17 at 15:41

Sumit S Chawla · Answer 3 · 2017-11-18T15:27:59.597

-1

Check the code below:

todolist = []

def add_item(item):
    global todolist
    todolist = todolist + [item]

def main():
    add_item(1)
    print(todolist)


if __name__ == '__main__':
    main()

edited Nov 18 '17 at 15:27

answered Nov 18 '17 at 13:25

Sumit S Chawla

3,180
1
14
33

You don't need the global statement if you use append. – Daniel Roseman Nov 18 '17 at 13:39
oh.. my bad. thanks for correcting. – Sumit S Chawla Nov 18 '17 at 13:44
This uses append and the question says trying to make a function that works like append. – Josh J Nov 18 '17 at 15:24

How Do I fix an UnboundLocalError in Python 3?

3 Answers3