Nonlocal keyword in Python - Step-by-Step Execution

Concept Flow - Nonlocal keyword

Define outer function

↓

Define inner function

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Inner function uses 'nonlocal'

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Modify outer variable

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Call inner function

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Outer variable changed

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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.