Higher-order function declaration in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a higher-order function that takes a function as a parameter.

Kotlin

fun operateOnNumber(x: Int, operation: (Int) -> Int): Int {
    return operation([1])
}

Drag options to blanks, or click blank then click option'

AInt

Bx

Coperation

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the function name instead of the argument.

Using a variable that is not defined in the function.

The parameter operation is a function that takes an Int and returns an Int. We call it with x inside the function.

2fill in blank

medium

Complete the code to declare a higher-order function that returns a function.

Kotlin

fun makeMultiplier(factor: Int): (Int) -> Int {
    return [1](number: Int) = number * factor
}

Drag options to blanks, or click blank then click option'

Afun

Bval

Clambda

Dreturn

Attempts:

3 left

💡 Hint

Common Mistakes

Using val instead of fun to declare the function.

Trying to return a lambda without the fun keyword.

The function makeMultiplier returns another function. Using fun declares an anonymous function to return.

3fill in blank

hard

Fix the error in the higher-order function declaration that takes two functions as parameters.

Kotlin

fun combineOperations(x: Int, op1: (Int) -> Int, op2: (Int) -> Int): Int {
    return op2(op1([1]))
}

Drag options to blanks, or click blank then click option'

Aop1

Bop2

Cx

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the function op1 itself instead of its argument.

Using an undefined variable instead of x.

The function combineOperations applies op1 to x, then op2 to the result. So x must be passed to op1.

4fill in blank

hard

Fill both blanks to declare a higher-order function that takes a function and returns a function.

Kotlin

fun transformFunction(f: (Int) -> Int): (Int) -> Int {
    return [1](x: Int) = f(x) + [2]
}

Drag options to blanks, or click blank then click option'

Afun

Bx

C1

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using val instead of fun.

Forgetting to add 1 to the result.

We declare an anonymous function with fun that takes x and returns f(x) + 1. The 1 is added to the result.

5fill in blank

hard

Fill all three blanks to declare a higher-order function that takes a function and returns a function that applies it twice.

Kotlin

fun twice(f: (Int) -> Int): (Int) -> Int {
    return [1](x: Int) = f(f([2])) + [3]
}

Drag options to blanks, or click blank then click option'

Afun

Bx

C0

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using val instead of fun.

Forgetting to pass x to f.

Adding a number other than 0.

The function twice returns an anonymous function declared with fun. It applies f twice to x and adds 0 (no change).

Practice

1. What is a higher-order function in Kotlin?

easy

A. A function that only returns integers

B. A function that takes another function as a parameter or returns a function

C. A function that cannot be called directly

D. A function that has no parameters

Solution

1. Step 1: Understand the definition of higher-order functions

   Higher-order functions are functions that can take other functions as parameters or return them as results.

2. Step 2: Match the definition with the options

   A function that takes another function as a parameter or returns a function correctly describes this behavior, while others describe unrelated function properties.

3. Final Answer:

   A function that takes another function as a parameter or returns a function -> Option B

4. Quick Check:

   Higher-order function = function with function parameter or return [OK]

Hint: Remember: higher-order means function uses another function [OK]

Common Mistakes:

• Thinking higher-order means function returns only numbers
• Confusing higher-order with functions without parameters
• Assuming higher-order functions cannot be called directly

2. Which of the following is the correct syntax to declare a higher-order function in Kotlin that takes a function parameter returning Int?

easy

A. fun operate(x: Int, f: Int): Int { return f(x) }

B. fun operate(x: Int, f: (Int) Int): Int { return f(x) }

C. fun operate(x: Int, f: (Int) -> Int): Int { return f(x) }

D. fun operate(x: Int, f: (Int) -> Int) { return f(x) }

Solution

1. Step 1: Identify correct function type syntax

   Kotlin uses (Type) -> ReturnType to declare function types. fun operate(x: Int, f: (Int) -> Int): Int { return f(x) } uses (Int) -> Int correctly.

2. Step 2: Check function return and parameter usage

   fun operate(x: Int, f: (Int) -> Int): Int { return f(x) } returns Int and calls f(x) properly. fun operate(x: Int, f: Int): Int { return f(x) } treats f as Int, not a function. fun operate(x: Int, f: (Int) Int): Int { return f(x) } misses arrow. fun operate(x: Int, f: (Int) -> Int) { return f(x) } lacks return type.

3. Final Answer:

   fun operate(x: Int, f: (Int) -> Int): Int { return f(x) } -> Option C

4. Quick Check:

   Function type syntax = (Int) -> Int [OK]

Hint: Look for (Type) -> ReturnType syntax for function parameters [OK]

Common Mistakes:

• Omitting the arrow '->' in function type
• Using wrong parameter type instead of function type
• Missing return type in function declaration

3. What is the output of this Kotlin code?

fun applyTwice(x: Int, f: (Int) -> Int): Int {
    return f(f(x))
}

fun increment(n: Int) = n + 1

fun main() {
    println(applyTwice(3, ::increment))
}

medium

A. 5

B. 4

C. 6

D. 3

Solution

1. Step 1: Understand applyTwice function behavior

   applyTwice calls the function f twice on x: f(f(x)). Here, f is increment which adds 1.

2. Step 2: Calculate the output step-by-step

   First increment(3) = 4, then increment(4) = 5, so applyTwice(3, increment) returns 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Increment twice on 3 = 5 [OK]

Hint: Apply the function twice stepwise to find output [OK]

Common Mistakes:

• Applying function only once
• Confusing function call syntax
• Ignoring the nested function call

4. Identify the error in this Kotlin higher-order function declaration:

fun transform(x: Int, f: Int -> Int): Int {
    return f(x)
}

medium

A. Missing parentheses around function type parameter

B. Return type of transform is missing

C. Function parameter name is invalid

D. Function body syntax is incorrect

Solution

1. Step 1: Check function type syntax in parameter

   Function types must be enclosed in parentheses like (Int) -> Int. Here, f: Int -> Int is missing parentheses.

2. Step 2: Verify other parts of the function

   Return type Int is present, parameter name f is valid, and function body syntax is correct.

3. Final Answer:

   Missing parentheses around function type parameter -> Option A

4. Quick Check:

   Function type needs parentheses: (Int) -> Int [OK]

Hint: Always wrap function types in parentheses (Type) -> ReturnType [OK]

Common Mistakes:

• Omitting parentheses around function type
• Confusing parameter name with type
• Missing return type declaration

5. How would you declare a Kotlin higher-order function named compose that takes two functions f and g (both from Int to Int) and returns a new function from Int to Int that applies g then f?

hard

A. fun compose(f: (Int) -> Int, g: (Int) -> Int): Int = f(g(0))

B. fun compose(f: Int, g: Int): Int = { x -> f(g(x)) }

C. fun compose(f: (Int) -> Int, g: (Int) -> Int): (Int) -> Int = f(g)

D. fun compose(f: (Int) -> Int, g: (Int) -> Int): (Int) -> Int = { x -> f(g(x)) }

Solution

1. Step 1: Understand function composition signature

   compose takes two functions f and g, both (Int) -> Int, and returns a function (Int) -> Int.

2. Step 2: Check the lambda expression for composition

   The returned function applies g first, then f: { x -> f(g(x)) }. fun compose(f: (Int) -> Int, g: (Int) -> Int): (Int) -> Int = { x -> f(g(x)) } matches this exactly.

3. Final Answer:

   fun compose(f: (Int) -> Int, g: (Int) -> Int): (Int) -> Int = { x -> f(g(x)) } -> Option D

4. Quick Check:

   Compose = return lambda applying g then f [OK]

Hint: Return a lambda applying g then f: { x -> f(g(x)) } [OK]

Common Mistakes:

• Returning wrong type instead of function
• Using incorrect parameter types
• Trying to call functions instead of returning a lambda