Nonlocal keyword in Python - Step-by-Step Execution

Concept Flow - Nonlocal keyword

Define outer function

↓

Define inner function

↓

Inner function uses 'nonlocal'

↓

Modify outer variable

↓

Call inner function

↓

Outer variable changed

↓

Return or print result

The flow shows how an inner function accesses and changes a variable from its outer function using 'nonlocal'.

Execution Sample

Python

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

This code changes the outer variable x from 5 to 10 using the nonlocal keyword inside inner().

Execution Table

Step	Action	Variable 'x' Value	Explanation
1	Define outer(), set x=5	5	Outer function starts, x initialized to 5
2	Define inner() inside outer()	5	Inner function created, no change yet
3	Call inner()	10	Before inner runs, x is 5; after inner runs, x is 10
4	Inside inner(), 'nonlocal x' declared	10	Inner function refers to outer x
5	Assign x = 10 inside inner()	10	Outer x changed to 10 via nonlocal
6	inner() ends, return to outer()	10	x remains 10 after inner finishes
7	outer() returns x	10	Final value of x is 10, changed by inner()

💡 inner() finishes, outer() returns updated x=10

Variable Tracker

Variable	Start	After Step 5	Final
x	5	10	10

Key Moments - 2 Insights

Why do we need 'nonlocal' inside inner() to change x?

Without 'nonlocal', assigning x inside inner() creates a new local x, not changing outer x. See step 4 and 5 where 'nonlocal x' links inner x to outer x.

What happens if we remove 'nonlocal' and assign x=10 inside inner()?

A new local variable x is created inside inner(), outer x stays 5. The execution_table would show x unchanged after inner() call.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of x right after step 3 (calling inner())?

A Undefined

B 5

C 10

D None

💡 Hint

Check the 'Variable x Value' column at step 3 in the execution_table.

At which step does the variable x change from 5 to 10?

A Step 4

B Step 5

C Step 6

D Step 7

💡 Hint

Look for the step where 'Assign x = 10 inside inner()' happens in the execution_table.

If we remove 'nonlocal' from inner(), what will outer() return?

A 5

B 10

C Error

D None

💡 Hint

Refer to key_moments about what happens without 'nonlocal' and the variable_tracker showing x values.

Concept Snapshot

Nonlocal keyword lets inner functions modify variables in outer (but not global) scopes.
Syntax: use 'nonlocal var' inside inner function.
Without it, assignment creates a new local variable.
Use it to change outer variables from inner functions.
Common in nested functions to share state.

Full Transcript

This visual trace shows how the 'nonlocal' keyword works in Python. We start by defining an outer function with a variable x set to 5. Inside it, we define an inner function. When inner() runs, it declares 'nonlocal x' to tell Python that x refers to the outer variable. Then inner() changes x to 10. After inner() finishes, the outer function returns x, which is now 10. Without 'nonlocal', inner() would create a new local x and the outer x would stay 5. This trace helps understand how 'nonlocal' allows inner functions to modify variables in their enclosing scopes.

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