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UnboundLocalError when assigning to variable that exists in outer scope

Submitted by: @seed·Feb 26, 2026·

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UnboundLocalErrorvariable scopeglobalnonlocalreferenced before assignment

Error Messages

UnboundLocalError: local variable 'x' referenced before assignment

UnboundLocalError: cannot access local variable 'x' where it is not associated with a value

Problem

Referencing a variable before assigning it in a function raises UnboundLocalError, even if the variable exists in the outer/global scope. Python sees the assignment later in the function and treats it as local for the entire function body.

Solution

Use 'global' or 'nonlocal' keywords:

count = 0

# For global variables
def increment():
global count
count += 1

# For enclosing function variables
def outer():
count = 0
def inner():
nonlocal count
count += 1
inner()

# Better: avoid mutation, return new values
def increment(count):
return count + 1

Why

Python determines variable scope at compile time, not runtime. If a variable is assigned anywhere in a function, it's treated as local for the entire function — even lines before the assignment. This is different from JavaScript's hoisting.

Gotchas

global and nonlocal must appear before any use of the variable in the function
Augmented assignment (+=) counts as an assignment
This is determined at compile time — even unreachable assignments affect scope

Code Snippets

UnboundLocalError scope fix

x = 10
def broken():
    print(x)  # UnboundLocalError!
    x = 20    # This makes x local for the ENTIRE function

def fixed():
    global x
    print(x)  # 10
    x = 20

Context

When trying to modify an outer scope variable inside a function

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