Functor <constant>

["Philip K.F. Hölzenspies"]

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 at botik.ru
>
>
>
>
>
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