Bruno Oliveira - 1 year ago 62

Scala Question

While working trough the exercises and concepts of the book: *Functional Programming in Scala* by Paul Chiusano and RĂșnar Bjarnason, I stumbled upon the exercise of writing my own function to reverse a list.

The "suggestion" the authors gave to motivate readers to learn more, was to see if we can write such a function using a fold.

My "non-fold" version is the following:

`def myrev(arr:List[Int]):List[Int] = if (arr.length > 0) { myrev(arr.tail) :+ arr.head } else arr`

However, can someone give me some pointers on how to at least start the logic to reverse a list via a fold?

I know that:

`List(1,2,3,4).foldLeft(0)({case(x,y)=>x})`

gives me the initial "seed" value, which is 0 above, and that:

`List(1,2,3,4).foldLeft(0)({case(x,y)=>y})`

gives me 4 which is the last element of the list.

So I would need to supply as function to foldLeft a sort of "identity" function that can give me an element in a certain position and maybe use it to reverse the list somehow, but I feel at a loss.

I didn't fully understood some Haskell code I found online and I want to "struggle" and try to get there on my own only with some pointers, instead of simply blindly copy some solution, so any help would be greatly appreciated.

Answer Source

```
scala> List(1, 2, 3, 4).foldLeft(List.empty[Int])((result, currentElem) => currentElem :: result)
res2: List[Int] = List(4, 3, 2, 1)
```