Subject: Re: Instances for Set of Functor, Traversable? Newsgroups: gmane.comp.lang.haskell.cafe Date: Tuesday 27th July 2010 08:17:57 UTC (over 7 years ago) Lennart Augustsson wrote: > Try to make Set an instance of Functor and you'll see why it isn't. > It's very annoying. And yet the very simple, and old solution works. http://okmij.org/ftp/Haskell/types.html#restricted-datatypes We just properly generalize Functor, so that all old functors are new functors. In addition, many more functors become possible, including Set. In general, we can have functors fmap' :: (C1 a, C2 b) => (a -> b) -> f a -> f b Incidentally, even an Integer may be considered a functor: we can define the fmap' operation fitting the above signature, where the constraint C1 a is a ~ Integer. Although the use of OverlappingInstances is not required, the extension leads to the nicest code; all old functors just work. {-# LANGUAGE MultiParamTypeClasses, FlexibleInstances #-} {-# LANGUAGE OverlappingInstances #-} module FunctorEx where import Control.Monad import Data.Set as S class Functor' f a b where fmap' :: (a -> b) -> f a -> f b -- The default instance: -- All ordinary Functors are also extended functors instance Functor f => Functor' f a b where fmap' = fmap -- Now define a functor for a set instance (Ord a, Ord b) => Functor' S.Set a b where fmap' = S.map -- Define a degenerate functor, for an integer newtype I a = I Integer deriving Show instance Functor' I Integer Integer where fmap' f (I x) = I $ f x -- tests -- Lists as functors test_l = fmap' (+10) [1,2,3,4] -- [11,12,13,14] -- Sets as functors test_s = fmap' (\x -> x `mod` 3) $ S.fromList [1,2,3,4] -- fromList [0,1,2] -- Integer as functor test_i = fmap' (* (6::Integer)) $ I 7 -- I 42 |
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