On Sat, 31 Oct 2009, b1g3ar5 wrote:
I'm trying:
instance Num b => Num (a -> b) where fromInteger = pure . Prelude.fromInteger negate = fmap Prelude.negate (+) = liftA2 (Prelude.+) (*) = liftA2 (Prelude.*) abs = fmap Prelude.abs signum = fmap Prelude.signum
but the compiler rejects it with:
src\Main.hs:24:9: Could not deduce (Show (a -> b), Eq (a -> b)) from the context (Num b) arising from the superclasses of an instance declaration at src\Main.hs:24:9-29 Possible fix: add (Show (a -> b), Eq (a -> b)) to the context of the instance declaration or add an instance declaration for (Show (a -> b), Eq (a -> b)) In the instance declaration for `Num (a -> b)'
Could someone please explain this to me?
I thought that it might be that it couldn't work out the functions necessary for (a->b) to be in the classes Show and Eq - so I tried adding definitions for == ans show, but it made no difference.
You have to define instances for Show and Eq, that is methods 'show' and (==), because the Num class has these classes as superclasses. This has been criticised a lot and is e.g. not the case in NumericPrelude. However, I would not seriously define a Num instance for functions: http://www.haskell.org/haskellwiki/Num_instance_for_functions