Enclosing scope in Python - Step-by-Step Execution

Concept Flow - Enclosing scope

Define outer function

↓

Define inner function

↓

Inner function accesses outer variable

↓

Call inner function

↓

Use outer variable from inner function

Shows how an inner function can use variables from its outer function's scope.

Execution Sample

Python

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

result = outer()

Defines an outer function with a variable x, an inner function that uses x, and calls inner to get result.

Execution Table

Step	Action	Variable x	Inner function return	Result variable
1	Call outer()	x not set	N/A	N/A
2	Set x = 10 in outer	10	N/A	N/A
3	Define inner() inside outer	10	N/A	N/A
4	Call inner()	10	x + 5 = 15	N/A
5	inner() returns 15	10	15	N/A
6	outer() returns 15	10	15	15
7	Assign result = outer()	10	15	15

💡 outer() finishes and returns inner() result, assigned to result

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	Final
x	undefined	10	10	10	10
Inner function return	N/A	N/A	15	15	15
result	undefined	undefined	undefined	15	15

Key Moments - 2 Insights

Why can inner() access variable x even though x is not defined inside inner()?

Because inner() is inside outer(), it can use variables from outer()'s scope as shown in step 4 of the execution_table.

What happens if outer() is called multiple times? Does x keep its value?

Each call to outer() creates a new x variable, so x is reset to 10 every time outer() runs, as seen in step 2.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of x when inner() is called at step 4?

A 10

B 5

C undefined

D 15

💡 Hint

Check the 'Variable x' column at step 4 in the execution_table.

At which step does the inner function return its value?

A Step 3

B Step 5

C Step 4

D Step 6

💡 Hint

Look at the 'Inner function return' column in the execution_table.

If we change x to 20 in outer(), what will be the inner() return value?

A 15

B 10

C 25

D 20

💡 Hint

Recall inner() returns x + 5; see variable_tracker for how x affects the return.

Concept Snapshot

Enclosing scope means an inner function can use variables from its outer function.
Syntax: define inner inside outer, inner accesses outer's variables.
The inner function remembers outer's variables when called.
Useful for organizing code and closures.
Example: inner() returns x + 5 where x is from outer().

Full Transcript

This example shows how an inner function can access a variable defined in its outer function. When outer() is called, it sets x to 10 and defines inner(). Calling inner() returns x + 5, which is 15. The result is returned by outer() and stored in result. The variable x is accessible inside inner() because of the enclosing scope. Each call to outer() creates a new x. This helps organize code and share data between nested functions.

Practice

1. What does enclosing scope mean in Python functions?

easy

A. Inner functions can access variables from outer functions

B. Variables inside a function are always global

C. Functions cannot access variables outside their own block

D. Only global variables can be used inside functions

Solution

1. Step 1: Understand function nesting

   In Python, functions can be defined inside other functions, creating an inner and outer function relationship.

2. Step 2: Access rules for variables

   Inner functions can use variables defined in their outer functions, which is called the enclosing scope.

3. Final Answer:

   Inner functions can access variables from outer functions -> Option A

4. Quick Check:

   Enclosing scope = inner function accesses outer variables [OK]

Hint: Inner functions see outer variables unless shadowed [OK]

Common Mistakes:

• Thinking all variables inside functions are global
• Believing inner functions cannot use outer variables
• Confusing local and global scopes

2. Which of the following is the correct syntax to modify an outer variable inside an inner function?

easy

A. Just assign the variable directly without any keyword

B. Use global keyword inside the inner function

C. Use nonlocal keyword inside the inner function

D. Use outer keyword inside the inner function

Solution

1. Step 1: Understand variable modification rules

   To change a variable from an outer (enclosing) function inside an inner function, Python requires the nonlocal keyword.

2. Step 2: Differentiate from global

   global is for variables at the module level, not enclosing functions.

3. Final Answer:

   Use nonlocal keyword inside the inner function -> Option C

4. Quick Check:

   Modify outer variable = use nonlocal [OK]

Hint: Change outer function vars with 'nonlocal' inside inner function [OK]

Common Mistakes:

• Using 'global' instead of 'nonlocal' for outer function variables
• Trying to assign without any keyword causing UnboundLocalError
• Using a non-existent 'outer' keyword

3. What is the output of this code?

def outer():
    x = 5
    def inner():
        return x + 3
    return inner()
print(outer())

medium

A. 3

B. 5

C. Error

D. 8

Solution

1. Step 1: Trace variable usage

   Variable x is defined in outer() as 5. The inner function returns x + 3, which is 8.

2. Step 2: Function call and return

   outer() calls inner() and returns its result, so print(outer()) prints 8.

3. Final Answer:

   8 -> Option D

4. Quick Check:

   Inner uses outer x=5, returns 5+3=8 [OK]

Hint: Inner function can read outer variables directly [OK]

Common Mistakes:

• Expecting an error due to variable scope
• Confusing inner return value with outer variable
• Thinking inner cannot access outer variables

4. What is wrong with this code?

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

medium

A. It will print 15

B. UnboundLocalError because inner tries to modify x without nonlocal

C. SyntaxError due to wrong indentation

D. NameError because x is not defined anywhere

Solution

1. Step 1: Analyze variable assignment in inner()

   Inside inner(), x = x + 5 tries to modify x. Python treats x as local but it is used before assignment.

2. Step 2: Understand error cause

   This causes an UnboundLocalError because x is referenced before assignment locally. The fix is to declare nonlocal x.

3. Final Answer:

   UnboundLocalError because inner tries to modify x without nonlocal -> Option B

4. Quick Check:

   Modify outer var inside inner needs nonlocal [OK]

Hint: Assigning outer var inside inner needs 'nonlocal' keyword [OK]

Common Mistakes:

• Assuming it prints 15 without error
• Confusing UnboundLocalError with NameError
• Ignoring the need for 'nonlocal' when modifying outer vars

5. Consider this code:

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

What will be the output?

hard

A. [1, 2, 3]

B. [0, 1, 2]

C. [1, 1, 1]

D. Error due to missing global keyword

Solution

1. Step 1: Understand closure with nonlocal

   The counter() function returns increment(), which modifies count using nonlocal. This keeps count persistent across calls.

2. Step 2: Trace calls to c()

   Each call to c() increases count by 1 and returns it. So outputs are 1, then 2, then 3.

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

   Closure with nonlocal increments count each call [OK]

Hint: Use nonlocal to keep and update outer variable state [OK]

Common Mistakes:

• Expecting count to reset each call
• Confusing nonlocal with global
• Thinking it will raise an error without global