Chad Scherrer wrote:
Are (a -> [b]) and [a -> b] isomorphic? I'm trying to construct a functionf :: (a -> [b]) -> [a -> b] that is the (at least one-sided) inverse of f' :: [a -> b] -> a -> [b] f' gs x = map ($ x) gs
Anything better than this? f g = [\x -> g x !! n | n <- [0..]] -Yitz _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
