How do I pass an anonymous function to map()? - kotlin

Question

I'm working through a little code challenge in Kotlin.
Write a function that nests the elements of a list one level deeper by repeating that
element inside a new list a given number of times.
Examples
(nest [:a :b :c] 2) ;=> ((:a :a) (:b :b) (:c :c))
(nest [] 10) ;=> ()
(nest [1 2 3 4] 1) ;=> ((1) (2) (3) (4))
(nest [1 2 3] 0) ;=> (()()())
I implemented nest like so:
fun <R> nest(list: List<R>, repeat: Int) =
list.map{
mem -> generateSequence(1) { it }
.map { mem }
.take(repeat)
.toList()
}
However, I'm wondering how I'd write the lambda as a separate anonymous function, which I'd pass to map, like...
val anonFunction = ???
list.map(anonFunction)
I'm not sure how this would work syntactically.

Answer 1

Curly braces is what starts a lambda function in Kotlin. The only thing you need to do to assign an arbitary lambda to a val, is to let Kotlin know what type it has.
This is probably the simplest way to do it. I am giving the compiler the type of the parameter, but letting type inference take care of the return type.
fun <R> nest(list: List<R>, repeat: Int): List<List<R>> {
val anonFunction = { mem: R ->
generateSequence(1) { it }
.map { mem }
.take(repeat)
.toList()
}
return list.map(anonFunction)
}
You could also do it by specifying the full function type of anonFunction by declaring it like this:
fun <R> nest(list: List<R>, repeat: Int): List<List<R>> {
val anonFunction: (R) -> List<R> = { mem ->
generateSequence(1) { it }
.map { mem }
.take(repeat)
.toList()
}
return list.map(anonFunction)
}
Side note:
A simpler implementation of your challenge could be this:
fun <R> nest(list: List<R>, repeat: Int) = list.map { elem ->
List(repeat) { elem }
}