Labeled break and continue in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of labeled break in nested loops

What is the output of this Kotlin code using a labeled break?

Kotlin

fun main() {
    outer@ for (i in 1..3) {
        for (j in 1..3) {
            if (i * j > 3) break@outer
            print("$i$j ")
        }
    }
}

A 11 12 13 21

B 11 12 21 22 23

C 11 12 13 21 22 23 31 32 33

D 11 12 21 22 31 32

💡 Hint

The labeled break exits the outer loop when the condition is met.

The code prints pairs i,j until i*j > 3. When i=2 and j=2, 2*2=4 > 3 triggers break@outer, stopping all loops. So output is 11 12 13 21.

❓ Predict Output

intermediate

Output of labeled continue in nested loops

What does this Kotlin code print when using a labeled continue?

Kotlin

fun main() {
    outer@ for (i in 1..2) {
        for (j in 1..3) {
            if (j == 2) continue@outer
            print("$i$j ")
        }
    }
}

A 11 21 22 23

B 11 21

C 11 12 13 21 22 23

D 11 13 21 23

💡 Hint

The labeled continue skips to the next iteration of the outer loop when j == 2.

When j == 2, continue@outer skips the rest of the inner loop and moves to next i. So for i=1, j=1 prints 11, j=2 triggers continue@outer skipping j=3. For i=2, j=1 prints 21, j=2 triggers continue@outer skipping j=3. So output is 11 21.

🔧 Debug

advanced

Identify the error with labeled break usage

What error does this Kotlin code produce?

Kotlin

fun main() {
    for (i in 1..3) {
        inner@ for (j in 1..3) {
            if (i == 2) break@outer
            print("$i$j ")
        }
    }
}

A Unresolved reference: outer

B Syntax error: Missing colon after label

C No error, prints 11 12 13 21 22 23 31 32 33

D Runtime exception: Label not found

💡 Hint

Check if the label used in break exists in the code.

The code tries to break@outer, but no label named 'outer' is defined. This causes a compile-time error: Unresolved reference: outer.

📝 Syntax

advanced

Which option correctly uses labeled continue?

Which of these Kotlin code snippets correctly uses labeled continue to skip the current iteration of the outer loop?

A outer@ for (i in 1..2) { for (j in 1..2) { if (j == 2) continue@ else print("$i$j ") } }

B outer@ for (i in 1..2) { for (j in 1..2) { if (j == 2) continue outer else print("$i$j ") } }

C outer@ for (i in 1..2) { for (j in 1..2) { if (j == 2) continue@outer else print("$i$j ") } }

D outer@ for (i in 1..2) { for (j in 1..2) { if (j == 2) continue@inner else print("$i$j ") } }

💡 Hint

Labeled continue syntax requires '@' after the label name.

Option C correctly uses 'continue@outer' to skip the current iteration of the outer loop. Option C misses '@', C uses incomplete label, D uses undefined label 'inner'.

🚀 Application

expert

Count iterations skipped by labeled continue

Given this Kotlin code, what value is printed by println(count)?

Kotlin

fun main() {
    var count = 0
    outer@ for (i in 1..3) {
        for (j in 1..3) {
            count++
            if (i + j == 4) continue@outer
        }
    }
    println(count)
}

A 8

B 7

C 9

D 6

💡 Hint

The continue@outer skips the rest of the inner loop and moves to next i when i+j == 4.

The pairs (i,j) are (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3). When i+j==4, pairs are (1,3),(2,2),(3,1). At these points, continue@outer skips remaining inner loop iterations and increments count only for executed count++ statements. Counting executed increments gives 6.

Practice

1. What does the break@label statement do in Kotlin nested loops?

easy

A. It stops the loop identified by the label immediately.

B. It skips the current iteration of the innermost loop.

C. It restarts the entire program from the beginning.

D. It pauses the loop until a condition is met.

Solution

1. Step 1: Understand labeled break usage

   The break@label statement is used to exit a specific loop marked by the label, not just the innermost loop.

2. Step 2: Compare with other options

   Options B and D describe continue or pause behavior, which is incorrect. It restarts the entire program from the beginning. is unrelated to loop control.

3. Final Answer:

   It stops the loop identified by the label immediately. -> Option A

4. Quick Check:

   break@label stops labeled loop [OK]

Hint: break@label exits the labeled loop immediately [OK]

Common Mistakes:

• Confusing break@label with continue@label
• Thinking break@label only stops innermost loop
• Assuming break@label pauses instead of stops

2. Which of the following is the correct syntax to continue the outer loop labeled outerLoop in Kotlin?

easy

A. continue#outerLoop

B. continue outerLoop

C. continue(outerLoop)

D. continue@outerLoop

Solution

1. Step 1: Recall Kotlin labeled continue syntax

   Kotlin uses the @ symbol to refer to labels with break or continue, so the correct syntax is continue@label.

2. Step 2: Eliminate incorrect options

   Options B, C, and D use invalid syntax for labels in Kotlin.

3. Final Answer:

   continue@outerLoop -> Option D

4. Quick Check:

   Use continue@label to continue labeled loop [OK]

Hint: Use @ symbol with label for continue or break [OK]

Common Mistakes:

• Omitting @ symbol before label
• Using parentheses or other symbols with label
• Confusing syntax with other languages

3. What is the output of the following Kotlin code?

outer@ for (i in 1..3) {
    inner@ for (j in 1..3) {
        if (i == 2 && j == 2) break@outer
        print("$i$j ")
    }
}

medium

A. 11 12 13 21 22

B. 11 12 13 21

C. 11 12 13 21 22 23

D. 11 12 13

Solution

1. Step 1: Trace nested loops and break condition

   The outer loop runs i=1 to 3, inner loop j=1 to 3. When i=2 and j=2, break@outer stops the outer loop entirely.

2. Step 2: List printed values before break

   For i=1, j=1..3 prints 11 12 13. For i=2, j=1 prints 21, then at j=2 break@outer stops all loops. So output is "11 12 13 21".

3. Final Answer:

   11 12 13 21 -> Option B

4. Quick Check:

   break@outer stops all loops at i=2,j=2 [OK]

Hint: break@label stops all loops at labeled point [OK]

Common Mistakes:

• Assuming break@outer only stops inner loop
• Including values after break condition
• Confusing break@outer with continue@outer

4. Identify the error in this Kotlin code snippet:

outer@ for (i in 1..2) {
    for (j in 1..2) {
        if (i == 1) continue outer
        println("i=$i, j=$j")
    }
}

medium

A. continue@outer should be used instead of continue outer.

B. continue cannot be used with a label on an inner loop.

C. Label 'outer' is not defined.

D. No error; code runs correctly.

Solution

1. Step 1: Check labeled continue syntax

   Kotlin requires the @ symbol when using labels with break or continue, so it must be continue@outer.

2. Step 2: Verify label definition and usage

   The label 'outer' is correctly defined, so Label 'outer' is not defined. is wrong. Using continue with label without @ causes syntax error.

3. Final Answer:

   continue@outer should be used instead of continue outer. -> Option A

4. Quick Check:

   Use continue@label syntax to avoid errors [OK]

Hint: Always use @ with label in continue or break [OK]

Common Mistakes:

• Omitting @ symbol in labeled continue
• Thinking label must be on inner loop
• Assuming no error when syntax is wrong

5. Given nested loops labeled outer and inner, how can you skip only the current iteration of the outer loop when a condition in the inner loop is met?

hard

A. Use break@outer inside inner loop.

B. Use continue without label inside inner loop.

C. Use continue@outer inside inner loop.

D. Use break without label inside inner loop.

Solution

1. Step 1: Understand labeled continue effect

   Using continue@outer skips the current iteration of the outer loop, moving to the next iteration of outer.

2. Step 2: Compare with break and unlabeled continue

   break@outer would exit the outer loop entirely, not just skip iteration. Unlabeled continue affects only inner loop.

3. Final Answer:

   Use continue@outer inside inner loop. -> Option C

4. Quick Check:

   continue@outer skips outer loop iteration [OK]

Hint: Use continue@label to skip labeled loop iteration [OK]

Common Mistakes:

• Using break@outer instead of continue@outer
• Using unlabeled continue to skip outer loop
• Confusing break and continue effects