I'm no expert, but it looks like the generalization of that would be some f that took a list: f :: [a] -> b so what you'd have is a fold, right? foldr1 :: (a -> a -> a) -> [a] -> a Best, Philip Neustrom On Sun, Aug 10, 2008 at 11:47 AM, Michael Feathers wrote:

I wrote this function the other day, and I was wondering if I'm missing something.. whether there is already a function or idiom around to do this.

unlist3 :: (a -> a -> a -> b) -> [a] -> b unlist3 f (x:y:z:xs) = f x y z

I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over.

Thanks,

Michael

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--- On Sun, 8/10/08, Michael Feathers wrote:

I wrote this function the other day, and I was wondering if I'm missing something.. whether there is already a function or idiom around to do this.

unlist3 :: (a -> a -> a -> b) -> [a] -> b unlist3 f (x:y:z:xs) = f x y z

I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over.

Well, there's always haskell's Swiss Army Knife, TH:

module Unlist(unlistN) where

import Language.Haskell.TH

unlistN n = do f <- newName "f" xs <- sequence (replicate n (newName "x")) lamE [varP f,(foldr ((flip infixP) '(:)) (wildP) (map varP xs))] (foldl appE (varE f) (map varE xs))

{-# OPTIONS_GHC -XTemplateHaskell #-} module UseUnlist where

import Unlist

f i0 i1 i2 i3 = i0 + i1 + i2 + i3

x = $(unlistN 4) f [1,2,3,4,5,6,7]

rcg