`list[2:5] = mlist` This python code assigns mlist...

# getting-started

[Vishal Rao]

list[2:5] = mlist
This python code assigns mlist into list at indices 2 to 5. I am not able to do the same in kotlin using inbuilt methods. Can anyone help me.

[Sam]

I can't think of one call that will do the whole thing, but you could use two sublist operations, like this:

list.subList(2, 5).let {
    it.clear()
    it.addAll(mList)
}

[Sam]

And since we're in Kotlin and we love immutability, here's a more idiomatic version, which returns a new list rather than mutating the existing one:

val newList = list.take(2) + mList + list.drop(5)

[ephemient]

this has fewer intermediates copies:

val newList = buildList {
    addAll(list.subList(0, 2))
    addAll(mList)
    addAll(list.subList(5, list.size))
}

with a small list it doesn't matter, but if you're splicing many parts together, this is better than repeated List::plus. also, depending on your problem, perhaps a different data structure would work better

[Ruckus]

You can define an extension

operator fun <T> MutableList<T>.set(
    slice: IntRange,
    values: List<T>,
) {
    val sub = subList(
        slice.start,
        slice.endInclusive,
    }
    sub.clear()
    sub.addAll(values)
}

// Use
list[2..5] = mList

[Sam]

That's clever! And actually feels pretty clean to me. However using 2..5 could be misleading, as it implies a closed range, whereas subList doesn't include the upper bound.

[Sam]

To match the OP, I'd change the implementation to val sub = subList(slice.start, slice.endExclusive) and then use list[2..<5] = mList

[Ruckus]

Agreed, I was just trying to match the syntax as close as possible. I would personally probably just use two args list[2, 5] = mList

[Vishal Rao]

Thanks guys, lot's of insight