14 Feb
2005
14 Feb
'05
10:35 a.m.
On Mon, Feb 14, 2005 at 03:55:01PM +0100, Lennart Augustsson wrote:
Any definition can be made point free if you have a complete combinator base at your disposal, e.g., S and K.
Haskell has K (called const), but lacks S. S could be defined as spread f g x = f x (g x)
Given that large set of Haskell prelude functions I would not be surprised if spread could already be defined point free in Haskell. :)
-- Lennart
I hope this won't be considered cheating... import Control.Monad.Reader k :: a -> b -> a k = return s :: (a -> r -> b) -> (a -> r) -> a -> b s = flip (>>=) . flip Greetings, Remi -- Nobody can be exactly like me. Even I have trouble doing it.