Global keyword in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code using global keyword?

Look at the code below. What will it print?

Python

x = 5

def change():
    global x
    x = 10

change()
print(x)

A 5

B NameError

C 10

D UnboundLocalError

💡 Hint

Think about what the global keyword does inside a function.

The global keyword tells Python to use the variable x defined outside the function. So when x is set to 10 inside change(), it changes the global x. The print statement then shows 10.

❓ Predict Output

intermediate

What error does this code raise?

What error will this code produce?

Python

count = 0

def increment():
    count += 1

increment()

A UnboundLocalError

B NameError

C SyntaxError

D No error

💡 Hint

Look at how count is used inside the function without global.

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

🔧 Debug

advanced

Why does this code print 0 instead of 1?

Fix the code so that print(x) outputs 1.

Python

x = 0

def set_x():
    x = 1

set_x()
print(x)

A Add <code>global x</code> inside <code>set_x()</code> before assignment

B Remove the assignment inside <code>set_x()</code>

C Change <code>x = 1</code> to <code>return 1</code> and assign outside

D Declare <code>x</code> as a parameter of <code>set_x()</code>

💡 Hint

Think about how to modify the global variable inside a function.

Without global x, the assignment x = 1 creates a new local variable x inside the function. The global x remains 0. Adding global x tells Python to use the global variable.

❓ Predict Output

advanced

What is the output of this nested function code?

What will this code print?

Python

count = 0

def outer():
    def inner():
        global count
        count += 1
    inner()

outer()
print(count)

A 0

B UnboundLocalError

C NameError

D 1

💡 Hint

Notice the use of global inside the inner function.

The global count inside inner() means it modifies the global count. So count increases from 0 to 1. The print shows 1.

🧠 Conceptual

expert

Which option causes an error due to incorrect global usage?

Which code snippet will raise an error because of wrong use of the global keyword?

A

def f():
    global b
    print(b)

b = 3
f()

B

def f():
    global c
    c += 1

f()

C

def f():
    global a
    a = 5

f()
print(a)

D

def f():
    global d
    d = d + 1

d = 0
f()

💡 Hint

Consider if the global variable is initialized before use.

Option B raises a NameError: name 'c' is not defined. The global c declaration is used, but c += 1 requires reading the global c first, which does not exist yet. Simple assignment would create it, but read-modify-write operations like += fail if undefined.

Option B: Reads existing global b, fine.

Option B: Assigns to undefined a, creating it globally, fine.

Option B: Reads existing d (initialized to 0) and updates it, fine.

Practice

1.

What does the global keyword do in Python?

easy

A. It creates a new local variable inside a function.

B. It allows a function to modify a variable defined outside it.

C. It deletes a variable from the global scope.

D. It imports a module globally.

Solution

1. Step 1: Understand variable scopes

   Variables defined outside functions are global, inside functions are local by default.

2. Step 2: Role of global keyword

   The global keyword tells Python to use the global variable inside the function, allowing modification.

3. Final Answer:

   It allows a function to modify a variable defined outside it. -> Option B

4. Quick Check:

   global lets function change outside variable [OK]

Hint: Global lets functions change outside variables directly [OK]

Common Mistakes:

• Thinking global creates new local variables
• Confusing global with import statements
• Assuming global deletes variables

2.

Which of the following is the correct way to modify a global variable count inside a function?

count = 0
def increment():
    ?
    count += 1

easy

A. def count

B. local count

C. global count

D. nonlocal count

Solution

1. Step 1: Identify the need to modify global variable

   The function wants to increase the global variable count.

2. Step 2: Use correct keyword to access global variable

   Using global count inside the function tells Python to use the global count, not create a local one.

3. Final Answer:

   global count -> Option C

4. Quick Check:

   Use 'global' to modify global variables inside functions [OK]

Hint: Use 'global' before variable to modify it inside function [OK]

Common Mistakes:

• Using 'local' which is not a Python keyword
• Using 'nonlocal' which applies to enclosing functions, not globals
• Trying to define variable with 'def'

3.

What is the output of this code?

value = 5
def change():
    global value
    value = 10
change()
print(value)

medium

A. 5

B. Error

C. None

D. 10

Solution

1. Step 1: Analyze the function change()

   The function declares value as global and sets it to 10.

2. Step 2: Effect of calling change()

   Calling change() updates the global value from 5 to 10.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   global lets function update outside variable [OK]

Hint: global changes outside variable, so print shows updated value [OK]

Common Mistakes:

• Thinking print shows original value 5
• Expecting an error without global keyword
• Confusing local and global scopes

4.

Find the error in this code and choose the fix:

counter = 0
def add():
    counter += 1
add()
print(counter)

medium

A. Add global counter inside add() before incrementing.

B. Change counter to a local variable inside add().

C. Remove the increment line counter += 1.

D. Define counter inside add() without global.

Solution

1. Step 1: Identify the error cause

   Trying to increment counter inside add() without declaring it global causes UnboundLocalError.

2. Step 2: Fix by declaring global variable

   Adding global counter inside add() tells Python to use the global counter variable.

3. Final Answer:

   Add global counter inside add() before incrementing. -> Option A

4. Quick Check:

   Declare global to modify outside variable inside function [OK]

Hint: Declare global before modifying global variable inside function [OK]

Common Mistakes:

• Ignoring the need for global declaration
• Trying to create local variable with same name
• Removing increment instead of fixing scope

5.

You want to count how many times a function is called using a global variable calls. Which code correctly updates calls each time track() runs?

calls = 0
def track():
    ?
    calls += 1

track()
track()
print(calls)

hard

A. global calls

B. nonlocal calls

C. local calls

D. calls = 0

Solution

1. Step 1: Understand the goal

   We want to update the global variable calls inside the function track().

2. Step 2: Use the correct keyword

   Using global calls inside track() allows incrementing the global calls variable.

3. Step 3: Confirm output

   Calling track() twice increments calls from 0 to 2, so print(calls) outputs 2.

4. Final Answer:

   global calls -> Option A

5. Quick Check:

   Use global to update global counter inside function [OK]

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

Common Mistakes:

• Using 'nonlocal' which is for nested functions
• Trying to assign local variable without global
• Resetting calls inside function