How about... *Main> bunch () [1..16] [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] *Main> bunch (8 :+: ()) [1..16] [[1,2,3,4,5,6,7,8],[9,10,11,12,13,14,15,16]] *Main> bunch (8 :+: 4 :+: ()) [1..16] [[[1,2,3,4],[5,6,7,8]],[[9,10,11,12],[13,14,15,16]]] *Main> bunch (8 :+: 4 :+: 2 :+: ()) [1..16] [[[[1,2],[3,4]],[[5,6],[7,8]]],[[[9,10],[11,12]],[[13,14],[15,16]]]] Here's the hack :)
{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances, TypeFamilies #-}
infixr 6 :+: data Bunch b = Int :+: b
class Bunchable b a where type Bunched b a :: * bunch :: b -> [a] -> Bunched b a
instance Bunchable () a where type Bunched () a = [a] bunch () = id
instance Bunchable b a => Bunchable (Bunch b) a where type Bunched (Bunch b) a = [Bunched b a] bunch (n :+: r) = map (bunch r) . simpleBunch n
simpleBunch :: Int -> [a] -> [[a]] simpleBunch _ [] = [] simpleBunch n as = let (c,cs) = splitAt n as in c:simpleBunch n cs
The key here is that 'Bunch' reflects its lenght on its type, but '[]' doesn't. It may be possible to keep using a list, but it would be very ugly, I guess. -- Felipe.