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'.