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Kotlinprogramming~3 mins

Why Labeled break and continue in Kotlin? - Purpose & Use Cases

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The Big Idea

What if you could instantly jump out of any loop you want, no matter how deep it is?

The Scenario

Imagine you are cooking multiple dishes at once, and you need to stop cooking one dish immediately when a certain ingredient is missing, but continue cooking the others. Doing this without clear signals can get confusing fast.

The Problem

Without labeled break and continue, you might have to write complicated checks and flags to control which loop to stop or continue. This makes your code messy, hard to read, and easy to make mistakes.

The Solution

Labeled break and continue let you clearly say which loop to stop or skip to the next iteration. This keeps your code clean and your intentions clear, just like using labels on your cooking pots to know exactly which one to handle.

Before vs After
✗ Before
for (i in 1..3) {
  for (j in 1..3) {
    if (someCondition) break
  }
}
✓ After
outer@ for (i in 1..3) {
  for (j in 1..3) {
    if (someCondition) break@outer
  }
}
What It Enables

You can precisely control complex nested loops, making your code easier to understand and maintain.

Real Life Example

When searching for a specific item in a grid, labeled break lets you stop searching immediately once found, without extra checks.

Key Takeaways

Labeled break and continue help manage nested loops clearly.

They prevent messy and error-prone manual checks.

They make your code cleaner and easier to follow.

Practice

(1/5)
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
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