Labeled break and continue in Kotlin - Step-by-Step Execution

Concept Flow - Labeled break and continue

Start Outer Loop

↓

Start Inner Loop

↓

Check Condition

↓

Labeled break or continue

↓

Exit or Continue Loops

↩Back to Outer Loop or Exit

The program runs nested loops. Inside, it checks a condition to decide whether to break or continue a specific loop using labels.

Execution Sample

Kotlin

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

This code uses a labeled break to exit the outer loop when i and j both equal 2.

Execution Table

Step	i	j	Condition (i==2 && j==2)	Action	Output
1	1	1	false	Print i=1, j=1	i=1, j=1
2	1	2	false	Print i=1, j=2	i=1, j=2
3	1	3	false	Print i=1, j=3	i=1, j=3
4	2	1	false	Print i=2, j=1	i=2, j=1
5	2	2	true	break@outer (exit outer loop)
6	-	-	-	Exit loops

💡 At step 5, condition is true, so break@outer exits the outer loop and stops all looping.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	Final
i	-	1	1	1	2	2	-
j	-	1	2	3	1	2	-

Key Moments - 2 Insights

Why does the program stop completely at step 5 instead of just breaking the inner loop?

Because the break uses the label 'outer', it exits the outer loop, not just the inner one. See step 5 in the execution_table where break@outer is used.

What would happen if we used 'continue@outer' instead of 'break@outer' at step 5?

The program would skip the rest of the current outer loop iteration and start the next one. It would not exit all loops. This changes the flow after step 5.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'j' at step 3?

A 2

B 1

C 3

D 4

💡 Hint

Check the 'j' column in the execution_table row for step 3.

At which step does the labeled break cause the loops to exit?

A Step 4

B Step 5

C Step 6

D Step 3

💡 Hint

Look at the 'Action' column in the execution_table to find where break@outer happens.

If the condition was never true, how many times would 'println' run?

A 9 times

B 6 times

C 3 times

D 0 times

💡 Hint

Consider the nested loops from 1 to 3 for both i and j, and check the output count in execution_table.

Concept Snapshot

Labeled break and continue in Kotlin:
- Use labels to specify which loop to break or continue.
- Syntax: label@ for loops, then break@label or continue@label.
- break@label exits the labeled loop immediately.
- continue@label skips to next iteration of labeled loop.
- Useful in nested loops to control flow precisely.

Full Transcript

This example shows how Kotlin uses labeled break to exit an outer loop from inside an inner loop. The outer loop runs i from 1 to 3, and the inner loop runs j from 1 to 3. When both i and j equal 2, the condition triggers break@outer, which stops all looping immediately. The execution table tracks each step, showing variable values and actions. The variable tracker shows how i and j change over time. Key moments clarify why labeled break exits the outer loop, not just the inner one. The quiz tests understanding of variable values and loop control flow. The concept snapshot summarizes how labeled break and continue work in Kotlin for nested loops.

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