[Haskell-cafe] Existential Types (I guess)

[Tom Davie]

Aside from Neil's point about rank-2 polymorphism, you can of course just
parameterise your NumHolder type...

data Num a => NumHolder a = NumHolder a

instance Show a => Show NumHolder a where
  show (NumHolder x) = show x

instance Functor NumHolder where
  fmap f (NumHolder a) = NumHolder (f a)

It depends what you want to do with your NumHolder though.  What is the
purpose of this type?

Bob

On Fri, Jan 22, 2010 at 11:31 AM, Ozgur Akgun <ozgurakgun at gmail.com> wrote:

> Dear Cafe,
>
> I can write and use the following,
>
> data IntHolder = IntHolder Integer
>
> instance Show IntHolder where
>     show (IntHolder n) = show n
>
> liftInt :: (Integer -> Integer) -> IntHolder -> IntHolder
> liftInt f (IntHolder c) = IntHolder (f c)
>
> But I cannot generalise it to *Num:*
>
> data NumHolder = forall a. Num a => NumHolder a
>
> instance Show NumHolder where
>     show (NumHolder n) = show n
>
> liftNum :: (Num a) => (a -> a) -> NumHolder -> NumHolder
> liftNum f (NumHolder c) = NumHolder (f c)
>
> The error message I get is the following:
>
>     Couldn't match expected type `a' against inferred type `a1'
>       `a' is a rigid type variable bound by
>           the type signature for `liftNum' at Lifts.hs:54:16
>       `a1' is a rigid type variable bound by
>            the constructor `NumHolder' at Lifts.hs:55:11
>     In the first argument of `f', namely `c'
>     In the first argument of `NumHolder', namely `(f c)'
>     In the expression: NumHolder (f c)
>
>
> Regards,
>
>
> --
> Ozgur Akgun
>
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