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.

Practice

1. What does the global keyword do inside a Python function?

easy

A. It deletes a global variable.

B. It creates a new local variable inside the function.

C. It makes the function run faster.

D. It allows the function to modify a variable defined outside the function.

Solution

1. Step 1: Understand the role of global keyword

   The global keyword tells Python that the variable inside the function refers to the variable defined outside (in global scope).

2. Step 2: Effect on variable modification

   Without global, assigning a value creates a new local variable. With global, it modifies the existing global variable.

3. Final Answer:

   It allows the function to modify a variable defined outside the function. -> Option D

4. Quick Check:

   global keyword modifies outer variable [OK]

Hint: global lets functions change outside variables [OK]

Common Mistakes:

• Thinking global creates a new local variable
• Assuming global deletes variables
• Believing global affects performance

2. Which of the following is the correct way to declare a global variable inside a function?

easy

A. def global x:

B. global = x

C. global x

D. x global

Solution

1. Step 1: Recall correct syntax for global declaration

   The correct syntax is to write the keyword global followed by the variable name, separated by a space.

2. Step 2: Check each option

   global x matches the correct syntax. Others are invalid Python syntax.

3. Final Answer:

   global x -> Option C

4. Quick Check:

   global keyword followed by variable name [OK]

Hint: Use 'global' then variable name to declare global inside function [OK]

Common Mistakes:

• Using '=' with global keyword
• Trying to define global like a function
• Placing global after variable name

3. What will be the output of this code?

count = 5
def increment():
    global count
    count += 1
increment()
print(count)

medium

A. 6

B. None

C. Error

D. 5

Solution

1. Step 1: Understand the global variable usage

   The variable count is defined globally with value 5. Inside increment(), global count allows modifying this global variable.

2. Step 2: Trace the function call and print

   The function adds 1 to count, changing it from 5 to 6. Then print(count) outputs 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   global lets function change count to 6 [OK]

Hint: global lets function update outside variable [OK]

Common Mistakes:

• Expecting original value 5 to print
• Thinking global causes error here
• Assuming function returns None

4. Find the error in this code:

total = 10
def add():
    total += 5
add()
print(total)

medium

A. Missing global declaration inside add()

B. Syntax error in function definition

C. total is not defined globally

D. print statement is incorrect

Solution

1. Step 1: Identify variable scope issue

   Inside add(), total += 5 tries to modify total. Without global total, Python treats total as local but it's used before assignment, causing an error.

2. Step 2: Fix by adding global declaration

   Adding global total inside add() tells Python to use the global total variable, fixing the error.

3. Final Answer:

   Missing global declaration inside add() -> Option A

4. Quick Check:

   Modify global variable needs global keyword [OK]

Hint: Add global keyword to modify global variable inside function [OK]

Common Mistakes:

• Ignoring need for global keyword
• Thinking total is local automatically
• Assuming no error occurs

5. You want to count how many times a function is called using a global variable. Which code correctly updates the global counter?

hard

A. calls = 0
def func():
calls += 1
func()
print(calls)

B. calls = 0
def func():
global calls
calls += 1
func()
print(calls)

C. calls = 0
def func():
calls = calls + 1
func()
print(calls)

D. calls = 0
def func():
global calls
calls = calls
func()
print(calls)

Solution

1. Step 1: Understand global variable modification

   To update a global variable inside a function, you must declare it with global inside the function.

2. Step 2: Analyze each option

   calls = 0 def func(): calls += 1 func() print(calls) tries to increment without global, causing error. calls = 0 def func(): global calls calls += 1 func() print(calls) correctly uses global calls and increments. calls = 0 def func(): calls = calls + 1 func() print(calls) tries to read and assign without global, causing UnboundLocalError. calls = 0 def func(): global calls calls = calls func() print(calls) declares global but does not increment.

3. Final Answer:

   calls = 0 def func(): global calls calls += 1 func() print(calls) -> Option B

4. Quick Check:

   global keyword needed to update global variable [OK]

Hint: Use global keyword to update global counter inside function [OK]

Common Mistakes:

• Forgetting global keyword causes UnboundLocalError
• Assigning without global causes UnboundLocalError when reading variable
• Declaring global but not updating variable