But that's not really a solution, since it doesn't make a Functor instance for Set; it makes a Functor' instance for Set. If you are willing to not be upwards compatible then, yes, there are solutions. I think the best bet for an upwards compatible solutions is the associated constraints, www.cs.kuleuven.be/~toms/Research/papers/constraint_families.pdf On Tue, Jul 27, 2010 at 10:17 AM, wrote:

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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