Nonlocal keyword in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Understanding nonlocal variable modification

What is the output of this code?

Python

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

result = outer()
print(result)

A 10

B 5

C UnboundLocalError

D None

💡 Hint

Remember that 'nonlocal' allows inner functions to modify variables in the nearest enclosing scope.

The 'nonlocal' keyword lets the inner function change the variable 'x' defined in the outer function. So after calling inner(), x becomes 10.

❓ Predict Output

intermediate

Nonlocal with multiple nested functions

What will be printed by this code?

Python

def a():
    x = 1
    def b():
        x = 2
        def c():
            nonlocal x
            x = 3
        c()
        return x
    return b()

print(a())

A 1

B UnboundLocalError

C 2

D 3

💡 Hint

The 'nonlocal' keyword affects the nearest enclosing scope, not the global or outermost.

Inside function c(), 'nonlocal x' refers to x in function b(). So c() changes b()'s x to 3. But b() returns x after c() runs, which is 3. However, b()'s x was initially 2 and changed to 3 by c(), so the return is 3.

🔧 Debug

advanced

Identify the error caused by missing nonlocal

What error does this code raise when run?

Python

def outer():
    count = 0
    def inner():
        count += 1
        return count
    return inner()

print(outer())

A UnboundLocalError

B TypeError

C SyntaxError

D 0

💡 Hint

Think about what happens when you try to change a variable from an outer scope without 'nonlocal'.

The inner function tries to increment 'count' but Python treats 'count' as a local variable because of the assignment. Since 'count' is used before assignment, it raises UnboundLocalError.

❓ Predict Output

advanced

Effect of nonlocal on mutable objects

What is the output of this code?

Python

def outer():
    data = [1, 2, 3]
    def inner():
        data.append(4)
    inner()
    return data

print(outer())

A [1, 2, 3]

B [1, 2, 3, 4]

C SyntaxError

D TypeError

💡 Hint

Check if 'nonlocal' can be used with mutable objects like lists.

The inner() function modifies the 'data' list from the outer() scope by appending 4. Since 'data' is a mutable object, it can be modified without declaring 'nonlocal'. So outer() returns [1, 2, 3, 4].

❓ Predict Output

expert

Complex nested nonlocal variable changes

What is the output of this code?

Python

def f():
    x = 0
    def g():
        nonlocal x
        x = 1
        def h():
            nonlocal x
            x = 2
        h()
        return x
    return g()

print(f())

A 1

B 0

C 2

D UnboundLocalError

💡 Hint

Follow how 'nonlocal' changes the same variable 'x' in nested functions.

Function g() declares 'nonlocal x' referring to f()'s x. It sets x=1. Then h() also declares 'nonlocal x' referring to the same x and sets x=2. So after h() runs, x is 2. g() returns x, so f() returns 2.