Raising exceptions in Python - Mini Project: Build & Apply

Raising Exceptions in Python

📖 Scenario: You are building a simple program that checks user input for age eligibility to enter a club. If the age is below the allowed limit, the program should raise an exception to stop the process and inform the user.

🎯 Goal: Create a program that raises an exception when the user's age is below 18, using Python's raise statement.

📋 What You'll Learn

Create a variable called age with a specific integer value.

Create a variable called minimum_age set to 18.

Use an if statement to check if age is less than minimum_age.

Raise a ValueError with the message 'Age is below the minimum allowed.' when the condition is true.

Print 'Access granted.' if the age is valid.

💡 Why This Matters

🌍 Real World

Programs often need to check if input values are valid and stop running if they are not. Raising exceptions helps catch errors early.

💼 Career

Understanding how to raise exceptions is important for writing safe and reliable code in software development jobs.

1

Create the age variable

Create a variable called age and set it to the integer 16.

Python

# Create a variable called age and set it to 16
# Your code here

Hint

Use age = 16 to store the user's age.

2

Set the minimum age

Create a variable called minimum_age and set it to the integer 18.

Python

age = 16
# Create a variable called minimum_age and set it to 18
# Your code here

Hint

Use minimum_age = 18 to set the allowed age limit.

3

Raise an exception if age is too low

Write an if statement that checks if age is less than minimum_age. Inside the if, use raise ValueError('Age is below the minimum allowed.') to stop the program with an error.

Python

age = 16
minimum_age = 18
# Add an if statement to check if age is less than minimum_age
# Inside the if, raise a ValueError with the message 'Age is below the minimum allowed.'
# Your code here

Hint

Use if age < minimum_age: and then raise ValueError('Age is below the minimum allowed.').

4

Print access granted if age is valid

After the if statement, write print('Access granted.') to show a success message when no exception is raised.

Python

age = 16
minimum_age = 18
if age < minimum_age:
    raise ValueError('Age is below the minimum allowed.')
# Print 'Access granted.' below
# Your code here

Hint

Use print('Access granted.') after the if block.

Note: Because age is 16, the program will raise an error and not print 'Access granted.'

Practice

1. What does the raise statement do in Python?

easy

A. It creates a new variable.

B. It prints a message to the screen.

C. It stops the program and signals an error.

D. It repeats a block of code.

Solution

1. Step 1: Understand the purpose of raise

   The raise statement is used to stop the program when an error or unexpected situation occurs.

2. Step 2: Compare options with raise behavior

   Only It stops the program and signals an error. correctly describes that raise stops the program and signals an error.

3. Final Answer:

   It stops the program and signals an error. -> Option C

4. Quick Check:

   raise = stop program on error [OK]

Hint: Remember: raise means stop and show error [OK]

Common Mistakes:

• Thinking raise prints messages
• Confusing raise with variable creation
• Assuming raise repeats code

2. Which of the following is the correct syntax to raise a ValueError with a message "Invalid input"?

easy

A. raise ValueError("Invalid input")

B. throw ValueError("Invalid input")

C. raise new ValueError("Invalid input")

D. error ValueError("Invalid input")

Solution

1. Step 1: Recall Python syntax for raising exceptions

   In Python, the correct way to raise an exception is using raise ExceptionType("message").

2. Step 2: Check each option

   raise ValueError("Invalid input") uses correct syntax. Options B, C, and D use invalid keywords or extra words not used in Python.

3. Final Answer:

   raise ValueError("Invalid input") -> Option A

4. Quick Check:

   raise + ExceptionType + message = correct syntax [OK]

Hint: Use 'raise ExceptionType("message")' exactly [OK]

Common Mistakes:

• Using 'throw' instead of 'raise'
• Adding 'new' keyword like other languages
• Using 'error' keyword which doesn't exist

3. What will be the output of this code?

def check_age(age):
    if age < 18:
        raise ValueError("Too young")
    return "Access granted"

try:
    print(check_age(16))
except ValueError as e:
    print(e)

medium

A. Too young

B. Access granted

C. ValueError exception not caught

D. No output

Solution

1. Step 1: Analyze function behavior with age 16

   Since 16 < 18, the function raises a ValueError with message "Too young".

2. Step 2: Check exception handling in try-except

   The exception is caught by the except block, which prints the error message "Too young".

3. Final Answer:

   Too young -> Option A

4. Quick Check:

   Exception message printed = Too young [OK]

Hint: Exception message prints if caught in except [OK]

Common Mistakes:

• Assuming function returns 'Access granted'
• Thinking exception crashes program
• Missing that except prints the error message

4. Find the error in this code snippet:

def divide(a, b):
    if b == 0:
        raise "Cannot divide by zero"
    return a / b

print(divide(10, 0))

medium

A. No error, code runs fine

B. raise must be followed by an exception instance, not a string

C. Division by zero is allowed in Python

D. Function divide should return None when b is zero

Solution

1. Step 1: Identify the raise statement usage

   The code uses raise "Cannot divide by zero", which raises a string, not an exception object.

2. Step 2: Understand correct raise syntax

   Python requires raising an exception instance, e.g., raise ValueError("Cannot divide by zero").

3. Final Answer:

   raise must be followed by an exception instance, not a string -> Option B

4. Quick Check:

   raise needs exception object, not string [OK]

Hint: Always raise an exception object, not a string [OK]

Common Mistakes:

• Raising strings instead of exceptions
• Ignoring that division by zero causes error
• Assuming code runs without error

5. You want to create a function check_score(score) that raises a ValueError if the score is not between 0 and 100 (inclusive). Which code correctly implements this?

hard

A. def check_score(score):
if score <= 0 or score >= 100:
raise ValueError("Score must be 0-100")
return True

B. def check_score(score):
if 0 < score < 100:
raise ValueError("Score must be 0-100")
return True

C. def check_score(score):
if score == 0 or score == 100:
raise ValueError("Score must be 0-100")
return True

D. def check_score(score):
if score < 0 or score > 100:
raise ValueError("Score must be 0-100")
return True

Solution

1. Step 1: Understand the valid score range

   Score must be between 0 and 100, including 0 and 100.

2. Step 2: Check each condition for raising ValueError

   def check_score(score): if score < 0 or score > 100: raise ValueError("Score must be 0-100") return True raises error if score is less than 0 or greater than 100, correctly allowing 0 and 100.

3. Step 3: Verify other options

   def check_score(score): if 0 < score < 100: raise ValueError("Score must be 0-100") return True raises error incorrectly for valid scores between 0 and 100. def check_score(score): if score <= 0 or score >= 100: raise ValueError("Score must be 0-100") return True excludes 0 and 100 incorrectly. def check_score(score): if score == 0 or score == 100: raise ValueError("Score must be 0-100") return True raises error only if score equals 0 or 100, which is wrong.

4. Final Answer:

   def check_score(score): if score < 0 or score > 100: raise ValueError("Score must be 0-100") return True -> Option D

5. Quick Check:

   Raise error outside 0-100 inclusive = def check_score(score): if score < 0 or score > 100: raise ValueError("Score must be 0-100") return True [OK]

Hint: Use 'if score < 0 or score > 100' to check range [OK]

Common Mistakes:

• Using wrong comparison operators
• Excluding valid boundary values
• Raising error inside valid range