CODEWITHSUNDEEP
kotlin
David Soroko
David Soroko

Reputation: 9106

Implementing an interface with parametrized function

Suppose I have an interface

interface A {
    fun  calculate(n: Int): Int
}

I'd like to implement A in a class B (say) so that the calculation is provided in B's constructor, something like:

class B (f : (Int) -> Int) : A {
    override fun  calculate(n: Int): Int  //...somehow assign f to calculate
}

Can this be done without making f a property of B?

Hmm... this works:

interface A {
    val calculate: (n: Int) -> Int
}

class B(f: (Int) -> Int) : A {
    override val calculate = f
}

Can someone explain the intended syntactic difference between fun and val here? I am aware of Function definition: fun vs val but would like to get a 'language level' understanding.

Upvotes: 1

Views: 108

Answers (2)

yole
yole

Reputation: 97328

The simplest way to implement this is:

class B(val f : (Int) -> Int) : A {
    override fun calculate(n: Int): Int = f(n)
}

Upvotes: 3

mfulton26
mfulton26

Reputation: 31264

Yes, this can be done without making f a property of B. e.g. By using class delegation to an object expression:

class B(f: (Int) -> Int) : A by object : A {
    override fun calculate(n: Int) = f(n)
}

Even so, a private val seems more appropriate for this simple example.

Upvotes: 1

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