Break and continue in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to exit the loop when the value reaches 5.

MATLAB

for i = 1:10
    if i == [1]
        break;
    end
end

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

A5

B0

C10

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing a number outside the loop range.

Using continue instead of break.

The break statement stops the loop when i equals 5.

2fill in blank

medium

Complete the code to skip printing the number 3 using continue.

MATLAB

for i = 1:5
    if i == [1]
        continue;
    end
    disp(i)
end

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

A0

B1

C5

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using break instead of continue.

Choosing a number not in the loop range.

The continue statement skips the rest of the loop body when i is 3, so 3 is not printed.

3fill in blank

hard

Fix the error in the code to stop the loop when the sum exceeds 10.

MATLAB

sumVal = 0;
for i = 1:10
    sumVal = sumVal + i;
    if sumVal [1] 10
        break;
    end
end

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

A==

B>

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or == which do not stop at the right time.

Not using a comparison operator.

The loop should stop when the sum is greater than 10, so use >.

4fill in blank

hard

Fill both blanks to create a loop that skips even numbers and stops at 7.

MATLAB

for i = 1:10
    if mod(i, 2) == [1]
        continue;
    end
    if i == [2]
        break;
    end
    disp(i)
end

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

A0

B1

C7

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 for even check.

Stopping at the wrong number.

mod(i, 2) == 0 checks for even numbers to skip, and i == 7 stops the loop.

5fill in blank

hard

Fill all three blanks to print only odd numbers less than 10 and stop at 9.

MATLAB

for i = 1:15
    if mod(i, [1]) == [2]
        disp(i)
        if i == [3]
            break;
        end
    end
end

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

A2

B1

C9

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of 1 for odd check.

Stopping at a number greater than 9.

The code checks for odd numbers with mod(i, 2) == 1 and stops when i == 9.

Practice

1. What does the break statement do inside a MATLAB loop?

easy

A. Stops the entire loop immediately and exits it.

B. Skips the current iteration and continues with the next one.

C. Restarts the loop from the beginning.

D. Pauses the loop until a condition is met.

Solution

1. Step 1: Understand the purpose of break

   The break statement is used to exit a loop immediately, no matter where it is in the loop.

2. Step 2: Compare with other loop controls

   Unlike continue, which skips only the current iteration, break stops the whole loop.

3. Final Answer:

   Stops the entire loop immediately and exits it. -> Option A

4. Quick Check:

   break = stop loop [OK]

Hint: Break stops loop fully; continue skips one step [OK]

Common Mistakes:

• Confusing break with continue
• Thinking break only skips one iteration
• Believing break restarts the loop

2. Which of the following is the correct syntax to skip the current iteration in a MATLAB for loop?

easy

A. continue;

B. next;

C. break;

D. skip;

Solution

1. Step 1: Identify the statement to skip iteration

   In MATLAB, continue is the keyword used to skip the rest of the current loop iteration and move to the next.

2. Step 2: Check other options

   break exits the loop, skip and next are not valid MATLAB keywords.

3. Final Answer:

   continue; -> Option A

4. Quick Check:

   continue = skip iteration [OK]

Hint: Use continue to skip current loop step [OK]

Common Mistakes:

• Using break instead of continue
• Writing skip or next which are invalid
• Omitting the semicolon (though optional)

3. What will be the output of this MATLAB code?

sum = 0;
for i = 1:5
    if i == 3
        break;
    end
    sum = sum + i;
end
sum

medium

A. 10

B. 3

C. 6

D. 15

Solution

1. Step 1: Trace the loop iterations

   The loop runs i = 1, 2, then at i = 3 the break triggers and loop stops.

2. Step 2: Calculate sum before break

   sum = 0 + 1 + 2 = 3; The addition at i=3 does not happen because break stops loop before sum update.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Sum before break = 3 [OK]

Hint: Break stops loop before adding i=3 [OK]

Common Mistakes:

• Adding i=3 before break
• Summing all numbers ignoring break
• Confusing break with continue

4. Find the error in this MATLAB code snippet:

for i = 1:5
    if i == 3
        continue
    end
    disp(i)
end

medium

A. Missing semicolon after continue

B. disp(i) will not display 3

C. No error, code runs correctly

D. continue cannot be used inside for loops

Solution

1. Step 1: Check syntax of continue

   In MATLAB, continue does not require a semicolon and is valid inside loops.

2. Step 2: Understand loop behavior

   When i == 3, continue skips disp(i), so 3 is not displayed. This is expected behavior.

3. Final Answer:

   No error, code runs correctly -> Option C

4. Quick Check:

   continue valid in loops [OK]

Hint: Continue works without semicolon in loops [OK]

Common Mistakes:

• Thinking semicolon is mandatory after continue
• Believing continue is invalid in for loops
• Expecting disp to print 3

5. You want to sum numbers from 1 to 10 but skip adding 5 and stop the loop if the sum exceeds 30. Which code correctly uses break and continue?

hard

A. sum = 0;
for i = 1:10
if sum > 30
continue;
elseif i == 5
continue;
end
sum = sum + i;
end

B. sum = 0;
for i = 1:10
if i == 5
break;
elseif sum > 30
continue;
end
sum = sum + i;
end

C. sum = 0;
for i = 1:10
if sum > 30
continue;
elseif i == 5
break;
end
sum = sum + i;
end

D. sum = 0;
for i = 1:10
if i == 5
continue;
end
sum = sum + i;
if sum > 30
break;
end
end

Solution

1. Step 1: Understand conditions order

   We must skip adding 5 (continue when i==5) and stop loop if sum > 30 (break after adding).

2. Step 2: Analyze each option

   sum = 0; for i = 1:10 if i == 5 continue; end sum = sum + i; if sum > 30 break; end end skips 5 before adding, then adds i, then checks sum > 30 to break. This matches requirements.

3. Final Answer:

   C -> Option D

4. Quick Check:

   Skip 5 with continue; break after sum check [OK]

Hint: Continue before add to skip; break after add to stop [OK]

Common Mistakes:

• Breaking when i==5 instead of continuing
• Checking sum > 30 before adding
• Using continue to skip after sum check