For–else execution behavior in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print "Found" if 3 is in the list, otherwise print "Not found".

Python

numbers = [1, 2, 3, 4, 5]
for num in numbers:
    if num == [1]:
        print("Found")
        break
else:
    print("Not found")

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

A3

B4

C5

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using a number not in the list, so the else block always runs.

Forgetting to use break to stop the loop when found.

The loop checks each number. When it finds 3, it prints "Found" and stops. The else runs only if the loop never breaks.

2fill in blank

medium

Complete the code to print "No even number" if no even number is found in the list.

Python

nums = [1, 3, 5, 7]
for n in nums:
    if n % [1] == 0:
        print("Even number found")
        break
else:
    print("No even number")

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

A3

B4

C5

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using a divisor other than 2, so the condition never matches.

Not using modulo operator correctly.

Even numbers are divisible by 2. The code checks if any number is divisible by 2. If none are, the else block runs.

3fill in blank

hard

Fix the error in the code to correctly print "Found" if 0 is in the list, else "Not found".

Python

values = [1, 2, 3, 4]
for val in values:
    if val == [1]:
        print("Found")
        break
else:
    print("Not found")

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

A0

B4

C5

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using a number that is actually in the list, which does not match the task.

Not understanding the else block runs only if break never happens.

The code looks for 0, but 0 is not in the list. Fixing the value to 0 makes the condition correct for the task.

4fill in blank

hard

Fill both blanks to create a dictionary of squares for numbers greater than 2.

Python

squares = {x: x[1]2 for x in range(1, 6) if x [2] 2}
print(squares)

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

A**

B>

C<

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication (*) instead of exponentiation (**).

Using less than (<) instead of greater than (>) in the condition.

The code uses exponentiation (**) to square numbers. The condition x > 2 filters numbers greater than 2.

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase keys and values greater than 1.

Python

result = [1]: [2] for k, v in data.items() if v [3] 1
print(result)

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

Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using k instead of k.upper() for keys.

Using < instead of > in the condition.

Swapping keys and values.

The keys are converted to uppercase with k.upper(). Values are kept as v. The condition filters values greater than 1.

Practice

1.

What happens to the else block in a for-else statement if the loop completes all iterations without a break?

easy

A. The else block runs after the loop finishes.

B. The else block never runs.

C. The else block runs before the loop starts.

D. The else block runs only if the loop has a break.

Solution

1. Step 1: Understand the for-else structure

   The else block in a for loop runs only if the loop finishes all iterations without encountering a break.

2. Step 2: Apply to the question

   If the loop completes normally, the else block executes after the loop ends.

3. Final Answer:

   The else block runs after the loop finishes. -> Option A

4. Quick Check:

   For-else else runs if no break [OK]

Hint: Else runs only if loop ends without break [OK]

Common Mistakes:

• Thinking else runs always
• Believing else runs before loop
• Confusing else with else-if

2.

Which of the following is the correct syntax for a for-else statement in Python?

___
    print(i)
else:
    print("Done")

easy

A. foreach i in range(3):

B. for i range(3):

C. for i in range(3):

D. for i to range(3):

Solution

1. Step 1: Recall correct for loop syntax

   Python uses for variable in iterable: syntax. So for i in range(3): is correct.

2. Step 2: Check options

   Only for i in range(3): matches correct syntax; others have syntax errors.

3. Final Answer:

   for i in range(3): -> Option C

4. Quick Check:

   Correct for loop syntax uses 'in' [OK]

Hint: Remember: for variable in iterable: [OK]

Common Mistakes:

• Omitting 'in' keyword
• Using 'to' instead of 'in'
• Using 'foreach' which is not Python

3.

What is the output of this code?

for num in [1, 2, 3]:
    if num == 4:
        break
else:
    print("No break encountered")

medium

A. No output

B. No break encountered

C. break

D. Error

Solution

1. Step 1: Analyze the loop and condition

   The loop checks if num == 4. Since 4 is not in the list, break never runs.

2. Step 2: Check the else block execution

   Because the loop completes without break, the else block runs and prints "No break encountered".

3. Final Answer:

   No break encountered -> Option B

4. Quick Check:

   Else runs if no break, so prints message [OK]

Hint: Else runs only if break never happens [OK]

Common Mistakes:

• Assuming else never runs
• Thinking break triggers else
• Expecting error due to missing break

4.

Find the error in this code snippet:

for i in range(3):
    if i == 1
        break
else:
    print("Loop ended")

medium

A. Missing colon after if condition

B. Missing colon after for loop

C. Indentation error in else block

D. break cannot be used inside for loop

Solution

1. Step 1: Check syntax of if statement

   The if statement is missing a colon at the end of the condition line.

2. Step 2: Verify other parts

   The for loop and else block have correct colons and indentation; break is valid inside loops.

3. Final Answer:

   Missing colon after if condition -> Option A

4. Quick Check:

   if statements require colon [:] [OK]

Hint: Check colons after if, for, else lines [OK]

Common Mistakes:

• Ignoring missing colon
• Thinking break is invalid
• Confusing indentation errors

5.

Consider this code:

def find_even(numbers):
    for n in numbers:
        if n % 2 == 0:
            print(f"Found even: {n}")
            break
    else:
        print("No even number found")

find_even([1, 3, 5])

What will be printed when calling find_even([1, 3, 5])?

hard

A. Found even: 3

B. Found even: 1

C. No output

D. No even number found

Solution

1. Step 1: Trace the loop with given list

   The list has only odd numbers: 1, 3, 5. None satisfy n % 2 == 0, so break never runs.

2. Step 2: Check else block execution

   Since no break occurred, the else block runs and prints "No even number found".

3. Final Answer:

   No even number found -> Option D

4. Quick Check:

   Else runs if no break, so prints no even found [OK]

Hint: Else runs only if no break in loop [OK]

Common Mistakes:

• Assuming else runs always
• Expecting break to run on odd numbers
• Ignoring else block behavior