Nonlocal keyword in Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the nonlocal keyword do in Python?

It allows a function to modify a variable defined in its nearest enclosing scope that is not global.

intermediate

Why can't you use global to modify variables in an enclosing function?

global only affects variables at the module (global) level, not variables inside enclosing functions. nonlocal is needed for enclosing function variables.

beginner

Given this code snippet, what will be printed?

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

It will print 10 because inner() changes x in the enclosing outer() function using nonlocal.

intermediate

What error occurs if you use nonlocal on a variable that doesn't exist in any enclosing scope?

Python raises a SyntaxError because nonlocal requires the variable to exist in an enclosing function scope.

beginner

Can nonlocal be used to modify global variables?

No. nonlocal only works with variables in enclosing functions. Use global to modify global variables.

What does the nonlocal keyword allow you to do?

A Modify a global variable

B Modify a variable in the nearest enclosing function scope

C Create a new local variable

D Delete a variable

nonlocal lets you change variables in the nearest enclosing function, not global or local variables.

What happens if you use nonlocal on a variable not found in any enclosing scope?

A Python raises a SyntaxError

B The variable is created locally

C The variable is created globally

D Nothing happens

Python requires the variable to exist in an enclosing function scope for nonlocal. Otherwise, it raises a SyntaxError.

Which keyword should you use to modify a global variable inside a function?

A enclosing

B local

C nonlocal

D global

global is used to modify variables at the module (global) level.

In this code, what will be printed?

def outer():
    x = 1
    def inner():
        x = 2
    inner()
    print(x)
outer()

A 2

B Error

C 1

D None

inner() creates a new local x, so outer()'s x remains 1.

Which of these is true about nonlocal?

A It can only be used inside nested functions

B It can be used anywhere in the program

C It creates a new global variable

D It deletes variables

nonlocal is only valid inside nested functions to access variables in the outer function.

Explain how the nonlocal keyword works with an example.

Think about nested functions and changing variables outside the inner function.

What is the difference between nonlocal and global keywords?

Consider where the variable lives: inside a function or at the module level.

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