I'm looking to generate a list of powers in Elm, up to a certain limit. I've written the following tests:

        [ test "works for negative limit" <| \_ -> equal (powers 2 -5) (Ok [])
        , test "works for 2 up to 63" <| \_ -> equal (powers 2 63) (Ok [1, 2, 4, 8, 16, 32])
        , test "works for 2 up to 64" <| \_ -> equal (powers 2 63) (Ok [1, 2, 4, 8, 16, 32])
        , test "works for 3 up to 81" <| \_ -> equal (powers 3 81) (Ok [1, 3, 9, 27, 81])
        , test "rejects negative bases" <| \_ -> equal (powers -2 63) (Err "negative base")

I have a solution that I'm not happy with because it seems pretty long. Now I know that when doing these kinds of things in the functional style, we generally use tail recursion (which is no problem). But as I am building up the list from small numbers to large numbers (e.g., 1 2 4 8 16 32 ...) I could keep appending to the accumulated list, but this is much slower than using prepend (::). So... I decided to prepend and then reverse at the very end:

powers base limit =
  if base < 0 then
    Err "negative base"
      helper base exponent limit soFar =
        let next = base ^ exponent in
          if next > limit then
            Ok soFar
            helper base (exponent + 1) limit (next :: soFar)
      case helper base 0 limit [] of
        Ok powerList -> Ok <| List.reverse powerList
        Err message -> Err message

This passes all the tests but I am wondering two things:

  1. Is consing (::) and then reversing at the end the right way to do this or is there a better way?
  2. Is there something amazing in Elm's List module that I missed that I should be using here?
  3. At the very end, when I wanted to reverse, was it necessary to do that matching? It seems unnecessarily verbose because I had to pass through the whole Err part. Since I only wanted to reverse a list in the Ok case and leave the Err part unchanged, was there a better way to write this?

1 Answer 1


Why don't you just use the exponentiation (^) operator? Using that you can omit the recursion at all:

powers : Int -> Int -> Result String (List Int)
powers base limit =
    if base < 0 then
        Err "negative base"
        let size = floor (logBase (toFloat base) (toFloat limit))
        in Ok (List.map (\i -> base ^ i) (List.range 0 size))

There are some functions I used here:

  • logBase calculates the logarithm of a number with a given base. This can help to calculate the size of the returned list.
  • List.range : Int -> Int -> List Int generates an Int list where every element increasing by one.
  • 1
    \$\begingroup\$ Ah beautiful. I was hoping to do it with List.map but did not think of List.range! \$\endgroup\$
    – Ray Toal
    Commented Nov 6, 2017 at 16:31

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