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.