Python's closure - local variable referenced before assignment

Question

I am trying to understand how closure works in Python.

I feel like add1 should work just fine here. I would expect the variable x to have been defined when helper is called. However, it is giving me a Local variable referenced before assignment error.

add2 is very similar to add1. Instead of assigning x with an integer, it assigns it with a dictionary. The behavior of it is also aligned with what I would expect. x is defined and reference-able inside helper.

import random

def add1():
    x = 0
    def helper():
        x = x + 1
        return x
    return helper

def add2():
    x = {}
    def helper():
        x[random.randint(1,1000)] = 3
        return x
    return helper

if __name__ == '__main__':
    a1 = add1()
    a2 = add2()

    # print(a1()) #This causes error
    print(a2()) #{650: 3}
    print(a2()) #{650: 3, 333: 3}

What's the logic behind this? What am I doing differently other than that the types of x are different?

Answer 1

You're expecting the compiler to know that the variable has been bound outside the closure. This is not true, and hence you need to use nonlocal to indicate this.

def add1():
    x = 0
    def helper():
        nonlocal x
        x = x + 1
        return x
    return helper

EDIT By denniss:

nonlocal is not necessary in add2 because it's just modifying x and not re-binding it (aka not re-assigning it). Whereas in add1, x= x+1 is a re-assignment.