Dear Serge, If you want the type checker to know what type is "inside" your BNatural, no there isn't. You use reverse in your example, which has type [a] -> [a], so the type of what is in your BNatural doesn't change when you fmap reverse onto it. In general, however, you can't guarantee this. What if you fmap length? fmap length (BNat [Bit0,Bit0]) -->> BNat 2 Obviously, this violates the declaration of BNat. In this case specific case, you could maybe reason that the type checker should just reduce that only [DBit] -> [DBit] functions are allowed. However, the Functor class is set up to be more general; it allows a -> b functions. When your data type can not be instantiated as Functor this way, then it "isn't a functor" (at least, in the Haskell interpretation of functor). There is nothing to stop you from defining your own alternative map: bmap :: ([DBit] -> [DBit]) -> BNatural -> BNatural bmap f (BNat bs) = BNat $ f bs I hope this makes some sense. Regards, Philip On Sep 9, 2009, at 6:45 PM, Serge D. Mechveliani wrote:
People,
I have data DBit = Bit0 | Bit1 deriving (Eq, Ord, Enum) data BNatural = BNat [DBit] deriving (Eq)
and want to apply things like fmap reverse (bn :: BNatural).
GHC reports an error on this usage of fmap. It also does not allow instance Functor BNatural where fmap f (BNat ds) = BNat $ f ds,
-- it reports that a type constructor for Functor must have kind * -
*. Probaly, GHC agrees with Haskell at this point (?).
Now, I deceive the compiler by declaring instead
newtype BNatAux a = BNat a deriving (Eq) type BNatural = BNatAux [DBit] instance Functor BNatAux where fmap f (BNat ds) = BNat $ f ds
So, a parasitic BNatAux is introduced. I wonder: is there a simpler way to have fmap for BNatural ? Thank you in advance for advice,
----------------- Serge Mechveliani mechvel@botik.ru
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