Enclosing scope in Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the enclosing scope in Python?

The enclosing scope is the scope of a function that surrounds another nested function. It is the middle layer between the local scope of the inner function and the global scope.

beginner

How does Python find a variable in nested functions?

Python looks for the variable first in the local scope, then in the enclosing scope (the outer function), then in the global scope, and finally in the built-in scope.

intermediate

What keyword allows a nested function to modify a variable in the enclosing scope?

The nonlocal keyword lets a nested function change a variable defined in its enclosing function's scope.

beginner

Example: What will this code print?

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

It will print 5 because the inner function accesses the variable x from the enclosing scope of outer.

intermediate

Why can't a nested function modify a variable in the enclosing scope without nonlocal?

Without nonlocal, assigning to a variable inside a nested function creates a new local variable, so the enclosing variable remains unchanged.

In Python, where does the enclosing scope refer to?

A The scope of the outer function surrounding a nested function

B The global scope of the program

C The local scope inside the nested function

D The built-in Python scope

The enclosing scope is the scope of the outer function that contains the nested function.

Which keyword allows a nested function to modify a variable in its enclosing scope?

A nonlocal

B global

C local

D enclose

The nonlocal keyword lets a nested function modify variables in the enclosing scope.

What happens if you assign a value to a variable inside a nested function without using nonlocal?

A It modifies the variable in the enclosing scope

B It creates a new local variable inside the nested function

C It causes a syntax error

D It modifies the global variable

Assigning without nonlocal creates a new local variable inside the nested function.

In the nested function, if a variable is not found locally, where does Python look next?

A Global scope

B Built-in scope

C It raises an error immediately

D Enclosing scope

Python looks in the enclosing scope after the local scope before checking global or built-in scopes.

What will this code print?

def outer():
    x = 10
    def inner():
        x = 20
        print(x)
    inner()
    print(x)
outer()

A 20 and 20

B 10 and 10

C 20 and 10

D 10 and 20

The inner function prints 20 (its local x), but the outer x remains 10.

Explain what the enclosing scope is and how Python uses it when looking up variables in nested functions.

Think about the layers Python checks for a variable name.

Describe how to modify a variable in an enclosing scope from inside a nested function and why this requires a special keyword.

Consider what happens when you assign a value inside a nested function.

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