Can we get a `MutableList<T>.removeLast(count: Int): List<T>`? :smile:

# stdlib

[Big Chungus]

Can we get a MutableList<T>.removeLast(count: Int): List<T>? 😄

[dmitriy.novozhilov]

mutableListOf(1, 2, 3).dropLast(n)

[Big Chungus]

Yeah, but drop doesn't actually remove items from the source list

[dmitriy.novozhilov]

Oh, I got what do you want

[Big Chungus]

It came up in AoC part 2. https://kotlinlang.slack.com/archives/C87V9MQFK/p1670216407003329

[ephemient]

.subList(...).clear()

[Big Chungus]

How is sublist relevant here?

[Big Chungus]

Doesn't it so return a new slice?

[Roukanken]

hmm, looks like subList is view of the list, via a backing field. Eg, any modifications on it, will reflect in parent list

val list = mutableListOf(0, 1, 2, 4, 5)
list.subList(1, 3).add(3)

println(list)  // [0, 1, 2, 3, 4, 5]

[Big Chungus]

Ok, looks like this is the code required to approximate removeLast(n)

val input = mutableListOf(1,2,3)
mutableListOf().also{ out ->
  input.subList(input.size - n, input.size).also(out.addAll).clear()
}

[ephemient]

surely this is easier?

val input = mutableListOf(1, 2, 3)
val slice = input.subList(input.size - n, input.size)
slice.toMutableList().also { slice.clear() }

[Big Chungus]

This is just semantics, the idea is the same.

[Dan Fingal-Surma]

What is the slice.toMutableList() part? That just creates a copy of the slice.

val input = mutableListOf(1, 2, 3)
	input.subList(input.size - 2, input.size).clear()
    println(input)

prints [1]

[Dan Fingal-Surma]

fun <T> MutableList<T>.removeLast(count: Int) {
    subList(size - count, size).clear()
}

[Roukanken]

Yeah, that copy is there on purpose: remove functions usually return what they removed. That last expression removes & returns a copy in one go