Raising exceptions in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to raise a ValueError with a message.

Python

def check_positive(number):
    if number <= 0:
        raise [1]("Number must be positive")

Drag options to blanks, or click blank then click option'

AValueError

Bprint

Creturn

DException

Attempts:

3 left

💡 Hint

Common Mistakes

Using print instead of raise

Using return instead of raise

Raising a generic Exception instead of ValueError

The raise keyword is used to throw an exception. Here, ValueError is the correct exception type to indicate a wrong value.

2fill in blank

medium

Complete the code to raise a custom exception with a message.

Python

class MyError(Exception):
    pass

def test():
    raise [1]("This is a custom error")

Drag options to blanks, or click blank then click option'

AMyError

BTypeError

CValueError

DException

Attempts:

3 left

💡 Hint

Common Mistakes

Raising a built-in exception instead of the custom one

Forgetting to define the custom exception class

Using the wrong exception class name

To raise a custom exception, use the class name defined for it. Here, MyError is the custom exception class.

3fill in blank

hard

Fix the error in raising an exception with a message.

Python

def check_age(age):
    if age < 18:
        raise ValueError[1]"Age must be at least 18"

Drag options to blanks, or click blank then click option'

A:

B,

C[

D(

Attempts:

3 left

💡 Hint

Common Mistakes

Using a comma instead of parentheses

Using a colon or bracket instead of parentheses

Not enclosing the message in quotes

When raising an exception with a message, the message must be inside parentheses after the exception type.

4fill in blank

hard

Fill both blanks to raise a TypeError with a custom message.

Python

def check_string(value):
    if not isinstance(value, str):
        raise [1]([2])

Drag options to blanks, or click blank then click option'

ATypeError

B"Value must be a string"

C"Invalid type"

DValueError

Attempts:

3 left

💡 Hint

Common Mistakes

Using ValueError instead of TypeError

Not putting the message inside parentheses

Using a variable instead of a string message

The raise statement needs the exception type and the message inside parentheses. Here, TypeError and a descriptive string message are correct.

5fill in blank

hard

Fill all three blanks to raise a RuntimeError with a formatted message including a variable.

Python

def check_status(status):
    if status != 'ok':
        raise [1](f"Status is [2] but expected [3]")

Drag options to blanks, or click blank then click option'

ARuntimeError

Bstatus

C'ok'

DValueError

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong exception type

Not using f-string syntax correctly

Forgetting to include the variable in the message

Use RuntimeError to raise the exception. The message uses an f-string to include the variable status and the expected value 'ok'.

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