Why scope matters in Python - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code with variable scope?

Look at the code below. What will it print?

Python

x = 10

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

func()
print(x)

A 10\n5

B 10\n10

C 5\n5

D 5\n10

💡 Hint

Remember, variables inside a function are local unless declared global.

The x inside func() is local and set to 5, so print(x) inside prints 5. Outside, x is still 10.

❓ Predict Output

intermediate

What error does this code raise due to scope?

What error will this code produce?

Python

def func():
    print(x)
    x = 3

func()

A UnboundLocalError

B NameError

C SyntaxError

D No error, prints 3

💡 Hint

Python treats variables assigned in a function as local, even if used before assignment.

Because x is assigned inside func(), Python treats it as local. The print(x) tries to access it before assignment, causing UnboundLocalError.

❓ Predict Output

advanced

What is the output when modifying a global variable inside a function?

What will this code print?

Python

count = 0

def increment():
    global count
    count += 1

increment()
increment()
print(count)

A 2

B 0

C NameError

D 1

💡 Hint

Using global allows modifying the variable outside the function.

The global keyword tells Python to use the global count. Each call to increment() adds 1, so after two calls, count is 2.

❓ Predict Output

advanced

What is the output of nested function scope?

What will this code print?

Python

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

outer()

A SyntaxError

B inner

C outer

D NameError

💡 Hint

nonlocal lets inner functions modify variables in the nearest enclosing scope.

The nonlocal keyword allows inner() to change x in outer(). So print(x) prints 'inner'.

🧠 Conceptual

expert

Why does this code produce an error related to scope?

Consider this code:

def f():
    print(a)
    a = 5
f()

Why does it raise an error?

A Because <code>a</code> is not defined anywhere globally

B Because <code>print</code> cannot be used inside functions

C Because <code>a</code> is used before assignment in the local scope

D Because Python does not allow variable assignment inside functions

💡 Hint

Think about how Python decides if a variable is local or global.

Python sees the assignment a = 5 inside f() and treats a as local. The print(a) tries to access this local a before it is assigned, causing an UnboundLocalError.