Custom error messages in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code with a custom error message?

Look at the code below. What will it print when run?

Python

def check_age(age):
    if age < 18:
        raise ValueError("Age must be at least 18")
    return "Access granted"

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

A Age must be at least 18

B Access granted

C ValueError

D TypeError

💡 Hint

The function raises an error if age is less than 18, and the error message is printed.

The function raises a ValueError with the message 'Age must be at least 18' when age is 16. The except block catches this and prints the message.

❓ Predict Output

intermediate

What error message is shown here?

What will this code print when run?

Python

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

try:
    print(divide(10, 0))
except ZeroDivisionError as e:
    print(e)

A Cannot divide by zero!

B None

C ZeroDivisionError

D division by zero

💡 Hint

The custom message inside the error is printed by the except block.

The function raises a ZeroDivisionError with the message 'Cannot divide by zero!' which is caught and printed.

❓ Predict Output

advanced

What is the output of this code with a custom exception class?

What will this code print when run?

Python

class MyError(Exception):
    def __init__(self, message):
        super().__init__(message)

try:
    raise MyError("Something went wrong!")
except MyError as e:
    print(f"Error: {e}")

A MyError

B Error: Something went wrong!

C Exception

D Something went wrong!

💡 Hint

The custom exception message is passed to the print statement with a prefix.

The custom exception MyError is raised with the message 'Something went wrong!'. The except block prints 'Error: ' plus the message.

❓ Predict Output

advanced

What error message is shown here with multiple exceptions?

What will this code print when run?

Python

def check_value(x):
    if x < 0:
        raise ValueError("Negative value not allowed")
    elif x == 0:
        raise ZeroDivisionError("Zero is not allowed")
    return x

try:
    check_value(0)
except ValueError as e:
    print(e)
except ZeroDivisionError as e:
    print(e)

A Negative value not allowed

B ZeroDivisionError

C ValueError

D Zero is not allowed

💡 Hint

The input is zero, so the ZeroDivisionError is raised and caught.

The function raises ZeroDivisionError with message 'Zero is not allowed' for input 0. The except block for ZeroDivisionError prints the message.

🧠 Conceptual

expert

Which option correctly defines a custom exception with a default message?

Choose the code that defines a custom exception class with a default error message that can be overridden.

A

class CustomError(Exception):
    message = "Default error message"
    def __init__(self, message):
        self.message = message

B

class CustomError(Exception):
    def __init__(self):
        self.message = "Default error message"
    def __str__(self):
        return self.message

C

class CustomError(Exception):
    def __init__(self, message="Default error message"):
        super().__init__(message)

D

class CustomError(Exception):
    def __init__(self, message):
        print(message)

💡 Hint

The correct class passes the message to the base Exception class and allows a default.

Option C correctly defines __init__ with a default message and calls super().__init__ to set it. This allows overriding the message or using the default.

Practice

1. What is the main purpose of using raise ValueError('Custom message') in Python?

easy

A. To print a warning message without stopping the program.

B. To stop the program and show a specific error message when a condition is not met.

C. To automatically fix errors in the code.

D. To ignore errors and continue running the program.

Solution

1. Step 1: Understand the raise statement

   The raise keyword is used to stop the program and throw an error.

2. Step 2: Purpose of custom messages

   Adding a message like 'Custom message' helps explain why the error happened.

3. Final Answer:

   To stop the program and show a specific error message when a condition is not met. -> Option B

4. Quick Check:

   raise with message = stop and explain error [OK]

Hint: Raise errors to stop and explain problems clearly [OK]

Common Mistakes:

• Thinking raise only prints messages without stopping
• Confusing raise with print or logging
• Believing raise fixes errors automatically

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

easy

A. raise 'Invalid input' ValueError

B. throw ValueError('Invalid input')

C. error ValueError('Invalid input')

D. raise ValueError('Invalid input')

Solution

1. Step 1: Identify the correct keyword

   In Python, raise is used to throw errors, not throw or error.

2. Step 2: Correct order of error and message

   The syntax is raise ErrorType('message'), so the error type comes first, then the message in parentheses.

3. Final Answer:

   raise ValueError('Invalid input') -> Option D

4. Quick Check:

   raise + ErrorType('message') = correct syntax [OK]

Hint: Use raise ErrorType('message') to create custom errors [OK]

Common Mistakes:

• Using throw instead of raise
• Placing message before error type
• Missing parentheses around the message

3. What will be the output of this code?

def check_age(age):
    if age < 18:
        raise ValueError('Age must be 18 or older')
    return 'Access granted'

print(check_age(16))

medium

A. Access granted

B. None

C. ValueError: Age must be 18 or older

D. SyntaxError

Solution

1. Step 1: Check the condition in the function

   The function raises a ValueError if age is less than 18. Here, age is 16, so the error triggers.

2. Step 2: Understand what happens on raise

   When the error is raised, the program stops and shows the error message instead of returning 'Access granted'.

3. Final Answer:

   ValueError: Age must be 18 or older -> Option C

4. Quick Check:

   raise triggers error output = ValueError message [OK]

Hint: If condition fails, raise stops and shows error message [OK]

Common Mistakes:

• Expecting function to return 'Access granted' anyway
• Confusing error message with print output
• Thinking raise prints message but continues

4. Find the error in this code snippet:

def check_number(num):
    if num < 0:
        raise 'Negative number error'
    return 'Number is positive'

print(check_number(-5))

medium

A. You cannot raise a string directly; it must be an Exception type.

B. The raise statement is missing parentheses.

C. The function should return None instead of a string.

D. The if condition should be num > 0.

Solution

1. Step 1: Check the raise statement

   The code tries to raise a string directly, which is not allowed in Python. Only Exception types can be raised.

2. Step 2: Correct way to raise errors

   Use raise ValueError('message') or another Exception class, not a plain string.

3. Final Answer:

   You cannot raise a string directly; it must be an Exception type. -> Option A

4. Quick Check:

   raise must use Exception type, not string [OK]

Hint: Always raise Exception objects, not strings [OK]

Common Mistakes:

• Raising strings instead of Exception classes
• Forgetting to include parentheses with message
• Changing condition incorrectly

5. You want to create a function validate_score(score) that raises a ValueError with the message "Score must be between 0 and 100" if the score is outside this range. Which code correctly implements this?

hard

A. def validate_score(score):
if score < 0 or score > 100:
raise ValueError('Score must be between 0 and 100')
return 'Valid score'

B. def validate_score(score):
if 0 <= score <= 100:
raise ValueError('Score must be between 0 and 100')
return 'Valid score'

C. def validate_score(score):
if score < 0 and score > 100:
raise ValueError('Score must be between 0 and 100')
return 'Valid score'

D. def validate_score(score):
if score == 0 or score == 100:
raise ValueError('Score must be between 0 and 100')
return 'Valid score'

Solution

1. Step 1: Understand the valid range condition

   The score is valid if it is between 0 and 100 inclusive. So invalid means less than 0 or greater than 100.

2. Step 2: Check the if condition logic

   def validate_score(score): if score < 0 or score > 100: raise ValueError('Score must be between 0 and 100') return 'Valid score' correctly uses if score < 0 or score > 100 to detect invalid scores and raise the error.

3. Final Answer:

   def validate_score(score): if score < 0 or score > 100: raise ValueError('Score must be between 0 and 100') return 'Valid score' -> Option A

4. Quick Check:

   Use or for invalid range, raise error if outside [OK]

Hint: Raise error if score is less than 0 or greater than 100 [OK]

Common Mistakes:

• Using and instead of or in condition
• Raising error for valid scores instead of invalid
• Checking only equality instead of range