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!
Aside from Neil's point about rank-2 polymorphism, you can of course just parameterise your NumHolder type...data Num a => NumHolder a = NumHolder ainstance Show a => Show NumHolder a whereshow (NumHolder x) = show xinstance Functor NumHolder wherefmap f (NumHolder a) = NumHolder (f a)It depends what you want to do with your NumHolder though. What is the purpose of this type?BobOn Fri, Jan 22, 2010 at 11:31 AM, Ozgur Akgun <ozgurakgun@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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