Kotlin: Passing operator as a function parameter

Question

I've the following function in Kotlin

fun evaluate(first:Int?, second:Int?) {
    var result = 0
    if (v.equals('*')) {
        result = (first ?: 0) * (second ?: 0)
    } else if (v.equals('+')) {
        result = (first ?: 0) + (second ?: 0)
    } else if (v.equals('-')) {
        result = (first ?: 0) - (second ?: 0)
    } else if (v.equals('/')) {
        result = (first ?: 0) / (second ?: 0)
    }
    return result   
}

I want to change it in a way so that I can pass as a third parameter needed operator and evalute the expression. Something like

fun evaluate(first:Int?, second:Int?, op: () -> Unit):Int {
    return (first ?: 0).op(second ?: 0)
}

How can I pass the operator as a function in this case? I checked the same kind of a question, but it is not clear how you can do that with operator.

Do you want to get rid of if-conditions? Where does the 'v' come from? — ilya.dorofeev, Sep 13 '16 at 12:36
@ilya.dorofeev hhe shows the final intended function that gets rid of all the if conditions. — Jayson Minard, Sep 13 '16 at 12:55

Jayson Minard · Accepted Answer · 2022-03-12T05:17:50.023

12

Writing a higher order function using a function type as a parameter allows using both built-in operators and lambda expressions for the operation, so this would look like:

fun evaluate(first: Int?, second: Int?, op: (Int, Int) -> Int): Int {
    return op(first ?: 0, second ?: 0)
}

Which can be called with built-in operators such as:

val r1 = evaluate(value1, value2, Int::times) 
val r2 = evaluate(value1, value2, Int::plus)
val r3 = evaluate(value1, value2, Int::minus) 
val r4 = evaluate(value1, value2, Int::div)

And with custom functions:

val r5 = evaluate(value1, value2) { a, b -> (a * a) + b }

Now you can also assign the operators to variables, such as you v:

val v: (Int, Int)->Int = Int::times  // typing needed on left to avoid ambiguous alternatives
// and then later...
val r6 = evaluate(value1, value2, v)

Note that a function written for signature Int.(Int)->Int can be passed into a parameter expecting (Int, Int)->Int because the receiver this will be passed in as the first parameter.

edited Mar 12 '22 at 05:17

answered Sep 13 '16 at 12:36

Jayson Minard

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1

This works as well: `evaluate2(value1, value2, Int::times)`, so you don't need a separate function, just the last one – voddan Sep 13 '16 at 16:28
Interesting that the signature doesn't match, but because of the receiver being the first parameter it still works. cool. Is that intentional or a bug? – Jayson Minard Sep 14 '16 at 13:29
confirmed it is intentional, wil update answer, thanks @voddan – Jayson Minard Sep 14 '16 at 13:33
It was a pre 1.0 feature) – voddan Sep 14 '16 at 14:44

score 1 · Answer 2 · answered Sep 13 '16 at 11:46

1

Change () -> Unit into Int.(Int) -> Int. Then all the other code should work as you've written it there.

On the calling side it this is the first int, and the first parameter is the second int: { other -> this * other }

answered Sep 13 '16 at 11:46

Kiskae

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score 0 · Answer 3 · answered Sep 16 '16 at 09:31

you can try do that:

fun evaluate(first: Int?, second: Int? , v:String ): Int = v.op(first ?: 0, second ?: 0)

fun String.op(first:Int,second:Int):Int = when (this) {
    "*" -> first * second
    "+" -> first + second
//....
    else -> throw Exception()
}       
fun main(args: Array<String>) {
    println(evaluate(2,3,"*"))
    println(evaluate(2,3,"+"))
}

Kotlin: Passing operator as a function parameter

3 Answers3