Closures and variable capture in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of closure capturing a loop variable

What is the output of this Kotlin code snippet?

Kotlin

fun main() {
    val funcs = mutableListOf<() -> Int>()
    for (i in 1..3) {
        funcs.add { i * 2 }
    }
    funcs.forEach { println(it()) }
}

A

2
4
6

B

6
6
6

C

2
2
2

D

1
2
3

💡 Hint

Remember that each lambda captures the current value of i in the loop.

Each lambda captures the current value of i during each iteration. So the outputs are 2, 4, and 6 respectively.

❓ Predict Output

intermediate

Closure capturing mutable variable

What will be printed by this Kotlin program?

Kotlin

fun main() {
    var x = 5
    val f = { x += 1; x }
    println(f())
    println(f())
}

A

6
7

B

5
6

C

7
8

D

6
6

💡 Hint

The lambda modifies the variable x each time it is called.

The lambda increments x by 1 each time it runs, so first it prints 6, then 7.

🔧 Debug

advanced

Why does this closure capture unexpected value?

This Kotlin code prints the same number three times instead of 1, 2, 3. Why?

Kotlin

fun main() {
    val funcs = mutableListOf<() -> Int>()
    var i = 1
    while (i <= 3) {
        funcs.add { i }
        i++
    }
    funcs.forEach { println(it()) }
}

A The variable i is captured by reference, so all lambdas see the final value 4.

B The lambdas capture a copy of i at each iteration, so output should be 1, 2, 3.

C The while loop does not increment i correctly, causing infinite loop.

D The list funcs is empty, so nothing is printed.

💡 Hint

Think about how the variable i is shared among all lambdas.

All lambdas capture the same variable i by reference. After the loop ends, i is 4, so all lambdas print 4.

❓ Predict Output

advanced

Output of nested closures with variable shadowing

What is the output of this Kotlin code?

Kotlin

fun main() {
    var x = 10
    val f = {
        val x = 5
        {
            x
        }
    }
    val g = f()
    println(g())
}

A 10

B 0

C Compilation error due to shadowing

D 5

💡 Hint

Inner lambda captures the closest x in scope.

The inner lambda captures the x declared inside f, which is 5, shadowing the outer x.

🧠 Conceptual

expert

Why does Kotlin capture variables by reference in closures?

Which is the main reason Kotlin closures capture variables by reference rather than by value?

A Because Kotlin does not support capturing variables in closures.

B To improve performance by copying variables each time the lambda is called.

C To allow lambdas to modify the original variable and reflect changes across calls.

D To prevent any changes to the variable after the lambda is created.

💡 Hint

Think about how closures can change variables outside their scope.

Kotlin captures variables by reference so that lambdas can modify and share the same variable state.