Break and continue behavior in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of break in a loop

What is the output of this Swift code?

Swift

var sum = 0
for i in 1...5 {
    if i == 3 {
        break
    }
    sum += i
}
print(sum)

A 10

B 6

C 15

D 3

💡 Hint

Remember, break stops the loop immediately.

The loop adds 1 and 2 to sum. When i is 3, break stops the loop, so sum is 1+2=3. The code adds sum += i after the if check. So for i=1, sum=1; i=2, sum=3; i=3 break happens before sum += i, so sum stays 3. So output is 3.

❓ Predict Output

intermediate

Output of continue in a loop

What is the output of this Swift code?

Swift

var sum = 0
for i in 1...5 {
    if i == 3 {
        continue
    }
    sum += i
}
print(sum)

A 15

B 10

C 12

D 9

💡 Hint

Continue skips the current loop iteration.

The loop adds all numbers from 1 to 5 except 3. So sum = 1+2+4+5 = 12.

🔧 Debug

advanced

Why does this break not stop the outer loop?

Consider this Swift code with nested loops. What is the output and why?

Swift

outerLoop: for i in 1...3 {
    for j in 1...3 {
        if j == 2 {
            break
        }
        print("i:\(i), j:\(j)")
    }
}

A

i:1, j:1
i:2, j:1
i:3, j:1

B

i:1, j:1
i:1, j:2
i:2, j:1
i:2, j:2
i:3, j:1
i:3, j:2

C i:1, j:1

D

i:1, j:1
i:2, j:1

💡 Hint

Break without label stops only the inner loop.

The break stops the inner loop when j == 2, so only j=1 prints for each i. Outer loop continues for i=1 to 3.

❓ Predict Output

advanced

Output with labeled continue

What is the output of this Swift code using labeled continue?

Swift

outer: for i in 1...3 {
    for j in 1...3 {
        if j == 2 {
            continue outer
        }
        print("i:\(i), j:\(j)")
    }
}

A i:1, j:1

B

i:1, j:1
i:2, j:1
i:3, j:1

C

i:1, j:1
i:1, j:3
i:2, j:1
i:2, j:3
i:3, j:1
i:3, j:3

D

i:1, j:1
i:1, j:2
i:2, j:1
i:2, j:2
i:3, j:1
i:3, j:2

💡 Hint

Labeled continue skips to next iteration of outer loop.

When j == 2, continue outer skips the rest of inner loop and goes to next i. So only j=1 prints for each i.

🧠 Conceptual

expert

Effect of break in switch inside loop

What is the output of this Swift code?

Swift

for i in 1...3 {
    switch i {
    case 2:
        break
    default:
        print(i)
    }
}

A

1
3

B

1
2
3

C 2

D 1

💡 Hint

Break inside switch exits switch, not the loop.

Break inside switch exits only the switch, so for i=2 nothing prints. For i=1 and i=3 default case prints i.

Practice

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

easy

A. Restarts the loop from the beginning.

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

C. Pauses the loop until a condition is met.

D. Stops the entire loop immediately and exits it.

Solution

1. Step 1: Understand the role of break in loops

   The break statement immediately stops the loop and exits it, no matter where it is in the loop.

2. Step 2: Compare with other options

   continue skips to the next iteration, but break stops the loop entirely. Restarting or pausing the loop is not the behavior of break.

3. Final Answer:

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

4. Quick Check:

   break stops loop [OK]

Hint: Break means stop the loop completely [OK]

Common Mistakes:

• Confusing break with continue
• Thinking break restarts the loop
• Assuming break pauses the loop

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

easy

A. skip

B. continue

C. break

D. stop

Solution

1. Step 1: Identify the keyword to skip iteration

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

2. Step 2: Verify other options

   break stops the loop, skip and stop are not valid Swift keywords for loops.

3. Final Answer:

   continue -> Option B

4. Quick Check:

   Skip iteration = continue [OK]

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

Common Mistakes:

• Using break instead of continue
• Using non-existent keywords like skip or stop
• Confusing continue with break

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

for i in 1...5 {
    if i == 3 {
        break
    }
    print(i)
}

medium

A. 1 2

B. No output

C. 3 4 5

D. 1 2 3 4 5

Solution

1. Step 1: Analyze the loop and break condition

   The loop runs from 1 to 5. When i equals 3, break stops the loop immediately.

2. Step 2: Determine printed values before break

   Values 1 and 2 are printed before reaching 3. At 3, loop stops, so 3, 4, 5 are not printed.

3. Final Answer:

   1 2 -> Option A

4. Quick Check:

   Break stops at 3, prints before it [OK]

Hint: Break stops loop before printing 3 [OK]

Common Mistakes:

• Assuming break skips only current iteration
• Thinking all numbers print
• Confusing break with continue

4. Find the error in this Swift code snippet:

for i in 1...5 {
    if i == 3 {
        continue
    }
    if i == 6 {
        break
    }
    print(i)
}

medium

A. break should come before continue in the code.

B. continue cannot be used inside a for loop.

C. The loop will never break because i never reaches 6.

D. The range 1...5 is invalid.

Solution

1. Step 1: Check the loop range and conditions

   The loop runs from 1 to 5, so i will never be 6.

2. Step 2: Analyze the break condition

   Since i == 6 never occurs, the break statement will never execute, so the loop runs fully except skipping 3 due to continue.

3. Final Answer:

   The loop will never break because i never reaches 6. -> Option C

4. Quick Check:

   Break condition unreachable [OK]

Hint: Check if break condition can ever be true [OK]

Common Mistakes:

• Thinking continue is invalid in for loops
• Assuming break runs before continue always
• Not checking loop range carefully

5. You want to print all numbers from 1 to 10 except 5 and stop printing after 8. Which code snippet correctly uses break and continue in Swift?

hard

A. for i in 1...10 {
if i == 5 { continue }
if i > 8 { break }
print(i)
}

B. for i in 1...10 {
if i == 5 { break }
if i > 8 { continue }
print(i)
}

C. for i in 1...10 {
if i == 5 { continue }
if i > 7 { continue }
print(i)
}

D. for i in 1...10 {
if i == 5 { break }
if i > 8 { break }
print(i)
}

Solution

1. Step 1: Understand requirements

   Skip printing 5 (use continue) and stop printing after 8 (use break when > 8).

2. Step 2: Check each option

   for i in 1...10 { if i == 5 { continue } if i > 8 { break } print(i) } uses continue to skip 5 and break to stop after 8, matching requirements exactly.

3. Final Answer:

   for i in 1...10 { if i == 5 { continue } if i > 8 { break } print(i) } -> Option A

4. Quick Check:

   Continue skips 5, break stops loop after 8 [OK]

Hint: Continue skips, break stops loop [OK]

Common Mistakes:

• Using break to skip 5 instead of continue
• Using continue to stop loop after 8
• Mixing order of break and continue incorrectly