Nonlocal keyword in Python

Introduction

The nonlocal keyword lets you change a variable in an outer function from inside a nested function. It helps when you want to keep track of changes without using global variables.

When you have a function inside another function and want to update a variable from the outer function.

When you want to remember a value that changes each time the inner function runs.

When you want to avoid using global variables but still share data between nested functions.

Syntax

Python

def outer_function():
    variable = None
    def inner_function():
        nonlocal variable
        variable = 'new_value'
    inner_function()

You must declare nonlocal before changing the variable inside the inner function.

The variable must exist in the nearest outer function scope, not global or local to inner function.

Examples

This example shows how nonlocal lets the inner function update count from the outer function.

Python

def counter():
    count = 0
    def increment():
        nonlocal count
        count += 1
        return count
    return increment

count_up = counter()
print(count_up())  # 1
print(count_up())  # 2

The inner function changes the message variable from the outer function using nonlocal.

Python

def greet():
    message = 'Hello'
    def change_message():
        nonlocal message
        message = 'Hi'
    change_message()
    print(message)

greet()  # Output: Hi

Sample Program

This program uses nonlocal to increase factor each time the inner function runs, changing the multiplication result.

Python

def make_multiplier():
    factor = 2
    def multiply(number):
        nonlocal factor
        factor += 1
        return number * factor
    return multiply

multiplier = make_multiplier()
print(multiplier(5))  # First call
print(multiplier(5))  # Second call

Important Notes

If you forget nonlocal, Python treats the variable as local and you get an error if you try to assign it.

nonlocal only works with variables in the nearest outer function, not global variables.

Summary

nonlocal lets inner functions change variables from outer functions.

It helps keep track of changing data without using global variables.

Always declare nonlocal before assigning to the variable inside the inner function.

Practice

1. What does the nonlocal keyword do in Python?

easy

A. Declares a variable as global across all modules.

B. Allows an inner function to modify a variable from its outer function.

C. Creates a new local variable inside the inner function.

D. Prevents any changes to variables in the outer function.

Solution

1. Step 1: Understand variable scopes

   Variables inside a function are local by default, and inner functions cannot change outer variables unless specified.

2. Step 2: Role of nonlocal

   The nonlocal keyword allows the inner function to access and modify variables from the nearest enclosing function scope.

3. Final Answer:

   Allows an inner function to modify a variable from its outer function. -> Option B

4. Quick Check:

   nonlocal changes outer function variable = A [OK]

Hint: Nonlocal lets inner functions change outer variables [OK]

Common Mistakes:

• Confusing nonlocal with global keyword
• Thinking nonlocal creates new variables
• Assuming nonlocal works outside functions

2. Which of the following is the correct syntax to use nonlocal inside a nested function?

easy

A. local variable_name

B. global variable_name

C. outer variable_name

D. nonlocal variable_name

Solution

1. Step 1: Recall the syntax for nonlocal

   The keyword nonlocal is followed by the variable name to indicate it refers to an outer function's variable.

2. Step 2: Compare options

   Only nonlocal variable_name uses the correct keyword and syntax: nonlocal variable_name.

3. Final Answer:

   nonlocal variable_name -> Option D

4. Quick Check:

   Correct nonlocal syntax = C [OK]

Hint: Use 'nonlocal' followed by variable name inside inner function [OK]

Common Mistakes:

• Using 'global' instead of 'nonlocal'
• Writing 'local' or 'outer' which are invalid keywords
• Forgetting to write variable name after nonlocal

3. What will be the output of the following code?

def outer():
    x = 5
    def inner():
        nonlocal x
        x = 10
    inner()
    return x
print(outer())

medium

A. 10

B. None

C. Error: no binding for nonlocal 'x'

D. 5

Solution

1. Step 1: Trace variable assignment

   Variable x is set to 5 in outer(). The inner function declares nonlocal x and sets x = 10.

2. Step 2: Effect of nonlocal on x

   The nonlocal keyword allows inner() to modify x in outer(). So after calling inner(), x becomes 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   nonlocal changes outer x to 10 = D [OK]

Hint: nonlocal lets inner change outer variable value [OK]

Common Mistakes:

• Thinking x remains 5 because inner is separate
• Expecting a syntax error for nonlocal usage
• Assuming inner creates a new local x

4. Find the error in this code snippet:

def counter():
    count = 0
    def increment():
        count = count + 1
        return count
    return increment()
print(counter())

medium

A. increment() should not return count

B. count should be global, not local

C. Missing 'nonlocal count' inside increment()

D. No error, code runs fine

Solution

1. Step 1: Identify variable scope issue

   Inside increment(), count = count + 1 tries to read and write count. Without nonlocal, Python treats count as local, but it is used before assignment.

2. Step 2: Fix with nonlocal

   Adding nonlocal count tells Python to use the count from counter(), allowing modification.

3. Final Answer:

   Missing 'nonlocal count' inside increment() -> Option C

4. Quick Check:

   Modify outer variable needs nonlocal = A [OK]

Hint: Add nonlocal before modifying outer variable inside inner function [OK]

Common Mistakes:

• Using global instead of nonlocal
• Ignoring variable scope causing UnboundLocalError
• Returning count without incrementing properly

5. Consider this code:

def make_accumulator():
    total = 0
    def add(value):
        nonlocal total
        total += value
        return total
    return add
acc = make_accumulator()
print(acc(5))
print(acc(3))
print(acc(-2))

What is the output of this code?

hard

A. 5\n8\n6

B. 5\n3\n-2

C. 0\n5\n8

D. Error: nonlocal used incorrectly

Solution

1. Step 1: Understand closure with nonlocal

   The function make_accumulator() returns add, which remembers total. The nonlocal total lets add update total each call.

2. Step 2: Trace calls to acc

   First call: total=0+5=5, prints 5.
   Second call: total=5+3=8, prints 8.
   Third call: total=8+(-2)=6, prints 6.

3. Final Answer:

   5 8 6 -> Option A

4. Quick Check:

   Accumulator sums values using nonlocal total = B [OK]

Hint: nonlocal keeps state in nested function closures [OK]

Common Mistakes:

• Expecting total to reset each call
• Confusing nonlocal with global
• Thinking output is separate values, not cumulative