Local scope in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of variable inside a function

What is the output of this code?

Python

def greet():
    message = "Hello"
    print(message)

greet()

A Hello

B None

C NameError

D message

💡 Hint

Look where the variable message is defined and where it is used.

The variable message is defined inside the function greet. When print(message) runs inside the function, it prints "Hello".

❓ Predict Output

intermediate

Accessing local variable outside function

What happens when you run this code?

Python

def show():
    number = 10

show()
print(number)

A None

B 10

C NameError

D 0

💡 Hint

Think about where number is defined and where you try to print it.

The variable number is local to the function show. It does not exist outside the function, so trying to print it outside causes a NameError.

❓ Predict Output

advanced

Local variable shadowing global variable

What is the output of this code?

Python

value = 5

def test():
    value = 10
    print(value)

test()
print(value)

A 10\n10

B 5\n10

C 5\n5

D 10\n5

💡 Hint

Check which value is printed inside and outside the function.

Inside the function, the local value is 10, so it prints 10. Outside, the global value is still 5, so it prints 5.

❓ Predict Output

advanced

Modifying global variable inside function without global keyword

What error or output does this code produce?

Python

count = 0

def increment():
    count += 1
    print(count)

increment()

A UnboundLocalError

B 1

C 0

D NameError

💡 Hint

Think about how Python treats variables assigned inside functions.

Because count is assigned inside the function, Python treats it as local. But it is used before assignment, causing an UnboundLocalError.

❓ Predict Output

expert

Understanding local scope with nested functions

Consider this code. What is the output when outer() is called?

Python

def outer():
    x = 'local'
    def inner():
        x = 'nonlocal'
        print(x)
    inner()
    print(x)

outer()

A local\nlocal

B nonlocal\nnonlocal

C local\nnonlocal

D nonlocal\nlocal

💡 Hint

Look at where x is assigned inside inner() and how it affects outer().

The inner() function assigns a new local x that shadows the x in outer(). So inner() prints 'nonlocal', but outer()'s x remains 'local'.

Practice

1. What does local scope mean in Python?

easy

A. Variables can be used anywhere in the program

B. Variables exist only inside the function where they are created

C. Variables are shared between all functions

D. Variables are stored permanently on the disk

Solution

1. Step 1: Understand the meaning of local scope

   Local scope means a variable is created inside a function and only exists there.

2. Step 2: Compare options with this meaning

   Only Variables exist only inside the function where they are created correctly states that variables exist only inside their function.

3. Final Answer:

   Variables exist only inside the function where they are created -> Option B

4. Quick Check:

   Local scope = variables inside function only [OK]

Hint: Local variables live only inside their function [OK]

Common Mistakes:

• Thinking local variables can be used outside the function
• Confusing local scope with global scope
• Believing variables are shared across functions

2. Which of the following is the correct way to define a local variable inside a function?

easy

A. def func():
x = 5

B. x = 5
def func():
print(x)

C. def func():
global x
x = 5

D. def func():
return x

Solution

1. Step 1: Identify local variable definition

   A local variable is created by assigning a value inside a function without global keyword.

2. Step 2: Check each option

   def func(): x = 5 assigns x inside the function, making it local. Others either use global or no assignment inside function.

3. Final Answer:

   def func():\n x = 5 -> Option A

4. Quick Check:

   Assign inside function = local variable [OK]

Hint: Assign variable inside function without global for local [OK]

Common Mistakes:

• Using global keyword when not needed
• Assigning variable outside function expecting it local
• Trying to return variable not defined inside function

3. What will be the output of this code?

def greet():
    message = "Hello"
    print(message)

greet()
print(message)

medium

A. Hello
NameError

B. Hello
Hello

C. NameError
Hello

D. NameError
NameError

Solution

1. Step 1: Understand variable scope in the code

   Variable 'message' is defined inside greet(), so it is local to that function.

2. Step 2: Trace the print statements

   Calling greet() prints 'Hello'. Then print(message) outside function causes NameError because 'message' is not defined globally.

3. Final Answer:

   Hello\nNameError -> Option A

4. Quick Check:

   Local variable outside function causes NameError [OK]

Hint: Local variables can't be printed outside their function [OK]

Common Mistakes:

• Assuming local variable is accessible globally
• Expecting both prints to show 'Hello'
• Ignoring NameError on second print

4. Find the error in this code and fix it:

def add():
    result = a + b
    print(result)

add()

Assuming a = 2 and b = 3 are defined outside the function.

medium

A. Add 'global a, b' inside add()

B. No error, code runs fine

C. Define a and b inside add()

D. Pass a and b as parameters to add()

Solution

1. Step 1: Identify variable scope issue

   a and b are defined outside but used inside add() without global or parameters, causing NameError.

2. Step 2: Fix by passing variables as parameters

   Passing a and b as parameters to add() allows access without global keyword.

3. Final Answer:

   Pass a and b as parameters to add() -> Option D

4. Quick Check:

   Use parameters to access outside variables inside function [OK]

Hint: Pass outside variables as parameters to use inside function [OK]

Common Mistakes:

• Using global keyword unnecessarily
• Defining variables inside function losing outside values
• Ignoring NameError from missing variables

5. How can you modify this code to keep track of how many times count_calls() is called, using local scope only?

def count_calls():
    calls = 0
    calls += 1
    print(f"Called {calls} times")

count_calls()
count_calls()

hard

A. Declare calls as global variable

B. Define calls outside function and modify inside

C. Use a default argument to store calls count

D. Use a class to store calls count

Solution

1. Step 1: Understand why calls resets

   Variable calls is local and resets to 0 each call, so count never increases.

2. Step 2: Use default argument to keep state locally

   Using a default argument like calls=[0] keeps count inside function without global or external variables.

3. Final Answer:

   Use a default argument to store calls count -> Option C

4. Quick Check:

   Default argument keeps local state across calls [OK]

Hint: Use default mutable argument to keep count inside function [OK]

Common Mistakes:

• Using global variable instead of local trick
• Defining calls outside function losing local scope
• Not realizing local variables reset each call