Handling specific exceptions in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of specific exception handling

What is the output of this code when the input is 0?

Python

try:
    result = 10 / 0
except ZeroDivisionError:
    print("Cannot divide by zero")
except Exception:
    print("Some other error")

A No output

B Some other error

C 0

D Cannot divide by zero

💡 Hint

Think about which exception is raised by dividing by zero.

Dividing by zero raises a ZeroDivisionError, so the first except block runs and prints "Cannot divide by zero".

❓ Predict Output

intermediate

Output when catching multiple exceptions

What will this code print?

Python

try:
    x = int('abc')
except (ValueError, TypeError):
    print("Caught ValueError or TypeError")
except Exception:
    print("Caught some other exception")

A Caught some other exception

B Caught ValueError or TypeError

C abc

D No output

💡 Hint

What exception does int('abc') raise?

int('abc') raises a ValueError, which is caught by the first except block.

❓ Predict Output

advanced

Exception handling with else and finally

What is the output of this code?

Python

try:
    print("Start")
    x = 5 / 1
except ZeroDivisionError:
    print("Divide by zero error")
else:
    print("No error occurred")
finally:
    print("Always runs")

A

Start
No error occurred
Always runs

B

Start
Divide by zero error
Always runs

C

No error occurred
Always runs

D

Start
Always runs

💡 Hint

Consider what happens when no exception is raised.

Since 5/1 does not raise an error, the else block runs after try, then finally always runs.

❓ Predict Output

advanced

What error is raised?

What error does this code raise?

Python

try:
    d = {'a': 1}
    print(d['b'])
except KeyError:
    print("Key not found")
except Exception:
    print("Other error")

A None

B Other error

C Key not found

D KeyError exception not caught

💡 Hint

What happens when you access a missing key in a dictionary?

Accessing d['b'] raises a KeyError, which is caught by the first except block.

❓ Predict Output

expert

Output with nested try-except and re-raising

What is the output of this code?

Python

try:
    try:
        x = 1 / 0
    except ZeroDivisionError:
        print("Inner catch")
        raise
except ZeroDivisionError:
    print("Outer catch")

A

Inner catch
Outer catch

B Inner catch

C Outer catch

D No output

💡 Hint

What happens when an exception is re-raised inside an except block?

The inner except catches and prints "Inner catch", then re-raises the exception. The outer except catches it and prints "Outer catch".

Practice

1. What is the main purpose of using try-except blocks in Python?

easy

A. To catch and handle specific errors so the program doesn't crash

B. To speed up the program execution

C. To write comments inside the code

D. To create new functions automatically

Solution

1. Step 1: Understand the role of try-except

   The try-except block is used to catch errors that happen during program execution.

2. Step 2: Identify the benefit of catching errors

   By catching errors, the program can handle them gracefully and continue running instead of crashing.

3. Final Answer:

   To catch and handle specific errors so the program doesn't crash -> Option A

4. Quick Check:

   try-except = catch errors [OK]

Hint: Try-except blocks catch errors to avoid crashes [OK]

Common Mistakes:

• Thinking try-except speeds up code
• Confusing try-except with comments
• Believing try-except creates functions

2. Which of the following is the correct syntax to catch a ZeroDivisionError in Python?

easy

A. try:
x = 1/0
except:
print('Error')

B. try:
x = 1/0
catch ZeroDivisionError:
print('Cannot divide by zero')

C. try:
x = 1/0
except ZeroDivisionError:
print('Cannot divide by zero')

D. try:
x = 1/0
except ZeroDivision:
print('Cannot divide by zero')

Solution

1. Step 1: Check the correct keyword for catching exceptions

   Python uses except to catch exceptions, not catch.

2. Step 2: Verify the exception name spelling

   The correct exception name is ZeroDivisionError, not ZeroDivision.

3. Final Answer:

   try: x = 1/0 except ZeroDivisionError: print('Cannot divide by zero') -> Option C

4. Quick Check:

   Use except + exact exception name [OK]

Hint: Use except with exact exception name to catch errors [OK]

Common Mistakes:

• Using 'catch' instead of 'except'
• Misspelling exception names
• Using generic except without specifying error

3. What will be the output of this code?

try:
    num = int('abc')
except ValueError:
    print('Value error caught')
except TypeError:
    print('Type error caught')

medium

A. Value error caught

B. Type error caught

C. No output

D. Program crashes with ValueError

Solution

1. Step 1: Identify the error raised by int('abc')

   Trying to convert 'abc' to int raises a ValueError.

2. Step 2: Match the error with except blocks

   The ValueError is caught by the first except block, so it prints 'Value error caught'.

3. Final Answer:

   Value error caught -> Option A

4. Quick Check:

   int('abc') = ValueError caught [OK]

Hint: Match error type to except block to find output [OK]

Common Mistakes:

• Confusing ValueError with TypeError
• Thinking program crashes without except
• Assuming no output if error caught

4. Find the error in this code and choose the correct fix:

try:
    print(10 / 0)
except ZeroDivisionError, e:
    print('Error:', e)

medium

A. Use except ZeroDivisionError(e):

B. Change except line to: except ZeroDivisionError as e:

C. Change print to print('Error') only

D. Remove the except block completely

Solution

1. Step 1: Identify the syntax error in except clause

   Python 3 requires 'as' to assign exception to a variable, not a comma.

2. Step 2: Correct the except syntax

   Replace except ZeroDivisionError, e: with except ZeroDivisionError as e:.

3. Final Answer:

   Change except line to: except ZeroDivisionError as e: -> Option B

4. Quick Check:

   Use 'as' to assign exception variable [OK]

Hint: Use 'except Exception as e:' syntax in Python 3 [OK]

Common Mistakes:

• Using comma instead of 'as' in except
• Removing except block causing crash
• Wrong parentheses in except clause

5. You want to handle both KeyError and IndexError in the same block. Which is the best way to write the except clause?

hard

A. except KeyError, IndexError:
print('Error caught')

B. except KeyError or IndexError:
print('Error caught')

C. except KeyError and IndexError:
print('Error caught')

D. except (KeyError, IndexError):
print('Error caught')

Solution

1. Step 1: Understand how to catch multiple exceptions

   Python requires a tuple of exceptions inside parentheses to catch multiple exceptions in one block.

2. Step 2: Identify correct tuple syntax

   The correct syntax is except (KeyError, IndexError): to catch both exceptions.

3. Final Answer:

   except (KeyError, IndexError): print('Error caught') -> Option D

4. Quick Check:

   Use tuple in except to catch multiple exceptions [OK]

Hint: Use except (Error1, Error2): to catch multiple exceptions [OK]

Common Mistakes:

• Using 'or' or 'and' instead of tuple
• Using comma without parentheses
• Trying to catch exceptions separately without blocks