Global scope in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of modifying a global variable inside a function

What is the output of this Python code?

Python

x = 10

def modify():
    global x
    x = 20

modify()
print(x)

A 20

B 10

C NameError

D None

💡 Hint

Think about how the 'global' keyword affects variable assignment inside functions.

The 'global' keyword tells Python to use the variable 'x' defined outside the function. So, when 'x' is assigned 20 inside 'modify()', it changes the global 'x'. Therefore, printing 'x' after calling 'modify()' outputs 20.

❓ Predict Output

intermediate

Output when modifying a variable without global keyword

What will this code print?

Python

count = 5

def increment():
    count = count + 1

increment()
print(count)

A UnboundLocalError

B 5

C 6

D NameError

💡 Hint

Consider what happens when you assign to a variable inside a function without declaring it global.

Inside 'increment()', Python treats 'count' as a local variable because it is assigned to. But it is used before assignment in 'count = count + 1', causing an UnboundLocalError.

🔧 Debug

advanced

Fix the code to update global list inside function

This code tries to add an item to a global list inside a function. Which option correctly fixes it?

Python

items = [1, 2, 3]

def add_item():
    items = items + [4]

add_item()
print(items)

A Change 'items = items + [4]' to 'items.append(4)' without global keyword.

B Change 'items = items + [4]' to 'items += [4]' without global keyword.

C Remove the function and add 4 directly to the list outside.

D Add 'global items' inside the function before modifying it.

💡 Hint

Remember that assigning to a variable inside a function creates a local variable unless declared global.

Option D correctly declares 'items' as global, so the assignment modifies the global list. Option D causes UnboundLocalError because 'items' is assigned inside the function without global. Option D avoids the function but does not fix the original code.

❓ Predict Output

advanced

Output of nested function modifying global variable

What does this code print?

Python

value = 1

def outer():
    def inner():
        global value
        value = 5
    inner()
    print(value)

outer()
print(value)

A 1\n5

B 1\n1

C 5\n5

D 5\n1

💡 Hint

The inner function uses 'global' to change the variable. Think about when the print statements run.

The inner function sets the global 'value' to 5. The print inside 'outer()' prints the global 'value' which is now 5. The final print also prints 5 because the global variable was changed.

🧠 Conceptual

expert

Understanding global variable behavior with immutable types

Given this code, what is the value of 'num' after calling 'change()'?

Python

num = 10

def change():
    global num
    num += 5

change()

A 10

B 15

C TypeError

D UnboundLocalError

💡 Hint

The '+=' operator modifies the global variable when declared global.

The 'global' keyword allows the function to modify the global variable 'num'. The '+=' operator adds 5 to the current value, so 'num' becomes 15.