Break and continue in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of break in a for loop

What is the output of this MATLAB code?

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

MATLAB

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

A

1
2
3

B

1
2

C

1
2
3
4
5

D

3
4
5

💡 Hint

The break statement stops the loop immediately when the condition is met.

The loop prints numbers starting from 1. When i equals 3, the break stops the loop before printing 3. So only 1 and 2 are printed.

❓ Predict Output

intermediate

Output of continue in a for loop

What is the output of this MATLAB code?

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

MATLAB

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

A

1
2
4
5

B

1
2
3
4

C 3

D

1
2
3
4
5

💡 Hint

The continue statement skips the current iteration when the condition is met.

When i equals 3, continue skips the disp(i) command, so 3 is not printed. All other numbers are printed.

🔧 Debug

advanced

Identify the error in break usage

What error does this MATLAB code produce?

if true
    break;
end

MATLAB

if true
    break;
end

A Logical error: infinite loop

B Runtime error: variable 'break' undefined

C Syntax error: 'break' not inside a loop

D No error, code runs fine

💡 Hint

Break can only be used inside loops in MATLAB.

The break statement must be inside a loop. Using it inside an if statement without a loop causes a syntax error.

❓ Predict Output

advanced

Nested loops with break and continue

What is the output of this MATLAB code?

for i = 1:3
    for j = 1:3
        if j == 2
            break;
        end
        fprintf('[%%d %%d]\n',i,j)
    end
end

MATLAB

for i = 1:3
    for j = 1:3
        if j == 2
            break;
        end
        fprintf('[%d %d]\n',i,j)
    end
end

A

[1 1]
[1 2]
[2 1]
[3 1]

B

[1 1]
[1 2]
[2 1]
[2 2]
[3 1]
[3 2]

C

[1 1]
[1 2]
[1 3]
[2 1]
[2 2]
[2 3]
[3 1]
[3 2]
[3 3]

D

[1 1]
[2 1]
[3 1]

💡 Hint

The break stops the inner loop when j equals 2, so only j=1 is displayed for each i.

For each i, the inner loop runs j=1 and prints [i 1]. When j=2, break stops the inner loop. So only [1 1], [2 1], and [3 1] are printed.

❓ Predict Output

expert

Complex loop control with break and continue

What is the output of this MATLAB code?

count = 0;
for i = 1:5
    if mod(i,2) == 0
        continue;
    end
    count = count + 1;
    if count == 3
        break;
    end
    disp(i)
end

MATLAB

count = 0;
for i = 1:5
    if mod(i,2) == 0
        continue;
    end
    count = count + 1;
    if count == 3
        break;
    end
    disp(i)
end

A

1
3

B

1
3
5

C

1
3
5
7

D

2
4

💡 Hint

Count only increases for odd numbers. When count reaches 3, break executes before disp(i).

The loop skips even numbers. For odd i, count increases. When count reaches 3 (at i=5), break stops the loop before printing 5. So only 1 and 3 are printed.

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