Re: [Haskell-cafe] Existential Types (I guess)

22 Jan 2010

      OK, I wasn't storing a simple (Num a) in my holder data structure, it was
more of a thing implementing one of my type classes. So it knows how to
quack, and I can use it now, thanks to Neil's suggestion.

I have a (somewhat) heterogeneus list of values, belonging to the same type
class, and I just wanted to map a function to that list.

Now it works.

BTW Tom, Thanks for your suggestion, but it simply doesn't fit my needs this
time. But I'll keep your suggestion in mind.

Cheers guys!

2010/1/22 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 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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Ozgur Akgun