Nonlocal keyword in Python - Mini Project: Build & Apply

Understanding the Nonlocal Keyword in Python

📖 Scenario: Imagine you are creating a simple counter inside a function. You want to increase the count each time you call a nested function. This project will help you learn how to use the nonlocal keyword to change a variable in the outer function from inside the inner function.

🎯 Goal: You will build a function with a nested function that increases a counter using the nonlocal keyword. You will then print the updated count.

📋 What You'll Learn

Create a function with a nested function

Use the nonlocal keyword to modify the outer variable

Call the nested function to increase the counter

Print the final counter value

💡 Why This Matters

🌍 Real World

Using <code>nonlocal</code> helps when you want to keep track of changes inside nested functions without using global variables. This is useful in creating counters, stateful functions, or closures.

💼 Career

Understanding <code>nonlocal</code> is important for writing clean, maintainable Python code, especially when working with nested functions, decorators, or advanced function designs.

1

Create the outer function with a counter variable

Create a function called counter that has a variable count set to 0 inside it.

Python

# Create the function counter with count = 0 inside
# Your code here

Hint

Define a function with def counter(): and inside it write count = 0.

2

Add a nested function that uses nonlocal to change count

Inside the counter function, create a nested function called increase. Use the nonlocal keyword for count inside increase. Then add 1 to count inside increase.

Python

def counter():
    count = 0
    # Create nested function increase
    # Use nonlocal count and add 1 to count
    # Your code here

Hint

Inside counter, define def increase():. Use nonlocal count to tell Python you want to change the outer count. Then write count += 1.

3

Call the nested function and return the count

Still inside counter, call the nested function increase() once. Then return the value of count from counter.

Python

def counter():
    count = 0
    def increase():
        nonlocal count
        count += 1
    # Call increase once
    # Return count
    # Your code here

Hint

Call increase() inside counter to add 1 to count. Then use return count to send the updated count back.

4

Print the result of calling counter

Call the function counter() and print its result using print(counter()).

Python

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

# Print the result of counter()
# Your code here

Hint

Use print(counter()) to show the number 1, which is the updated count.

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