Create copy of generic array with new size

Question

Is it possible in Kotlin to create a copy of a generic array with new size if I already have an instance of that array and pass a construction method for its items? I think about something like this:

fun <T> resize(arr: Array<T>, newSize: Int, creator: (Int) -> T): Array<T> {
    ...
}

Obviously I cannot call Array<T>(newSize) { i -> creator(i) } because the type T is not known at compile time. For efficicy reasons I do not want to use an inline function with a reified type. I also cannot use the arr.copyOf(newSize) method here because that would return an Array<T?>.

In Java I could use Arrays.copyOf(arr, newSize) even with a generic array because we don't have null safety here. But would this work in Kotlin as well? Or do you have any other ideas what I could do?

"For efficicy reasons I do not want to use an inline function with a reified type." Why? It should be more efficient by inlining creator. — Alexey Romanov
– Alexey Romanov, Commented Dec 5, 2018 at 7:35

Kevin Coppock · Accepted Answer · 2018-12-04 20:14:02Z

3

I would just add an extension method to Array<T> for this. You can still rely on Arrays.copyOf under the hood, but before returning the value fill any remaining spaces with the result of the creator block:

fun <T> Array<T>.resize(newSize: Int, creator: (Int) -> T): Array<T> {
    val copiedArray = Arrays.copyOf(this, newSize)
    for (i in size until newSize) { copiedArray[i] = creator(i) }
    return copiedArray
}

For example:

val array = arrayOf("a", "b", "c")

// same: [a, b, c]
println("same: ${Arrays.toString(array.resize(3) { it.toString() })}")

// less: [a, b]
println("less: ${Arrays.toString(array.resize(2) { it.toString() })}")

// more: [a, b, c, 3, 4]
println("more: ${Arrays.toString(array.resize(5) { it.toString() })}")

answered Dec 4, 2018 at 20:14

Kevin Coppock

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3 Comments

Cilenco Over a year ago

Thanks that worked great :) Don't know Kotlin does not provide this by default

Cilenco Over a year ago

But this is only possible because I already have an instance of that array right? I could not use this method to create a completely new array of a generic type right?

Kevin Coppock Over a year ago

Correct, but you can use the built-in function that @Sergey usedbelow for that -- e.g. Array(size) { index -> index.toString() }

Sergio · Accepted Answer · 2018-12-04 20:11:38Z

0

I guess this will work:

fun <T> resize(arr: Array<T>, dest: Array<T>, creator: (Int) -> T): Array<T> {
    val list = arrayListOf<T>()
    // fill the list
    for (i in 0 until dest.size) {
        list.add(creator(i))
    }
    return list.toArray(dest)
}

fun callResize() {
    val arr = Array(5) { " " }
    val dest = Array(10) { "" }
    val creator = { i: Int -> "$i"}
    resize(arr, dest, creator)
}

The result is in the dest array.

edited Dec 4, 2018 at 20:11

answered Dec 4, 2018 at 20:06

Sergio

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