Re: [Haskell-cafe] Dynamic and equality

You can define: data EqDyn= forall a.(Typeable a, Eq a)=> EqDyn a instance Eq EqDyn where (EqDyn x) == (EqDyn y)= typeOf x== typeOf y && x== unsafeCoerce y unsafeCoerce is safe synce the expression assures that types are equal 2013/7/20 adam vogt

On Sat, Jul 20, 2013 at 12:31 AM, Carter Schonwald wrote:

...

the tricky part then is to add support for other types.

another approach to existentially package type classes with the data type!

eg data HasEq = forall a . HasEq ( Eq a => a) or its siblinng data HasEq a = Haseq (Eq a => a )

note this requires more planning in how you structure your program, but is a much more pleasant approach than using dynamic when you can get it to suite your application needs.

note its also late, so I've not type checked these examples ;)

Hi Carter,

It doesn't seem like the existential one will work as-is, since ghc rejects this:

{-# LANGUAGE ExistentialQuantification #-} data HEQ = forall a. Eq a => HEQ a usingHEQ :: HEQ -> HEQ -> Bool usingHEQ (HEQ a) (HEQ b) = a == b

I think you were hinting at this option which is better than my first suggestion:

{-# LANGUAGE ExistentialQuantification #-} import Data.Typeable data DYN = forall a. Typeable a => DYN (a, DYN -> Bool)

mkDyn :: (Eq a, Typeable a) => a -> DYN mkDyn x = DYN (x, \(DYN (y, eq2)) -> case cast y of Just y' -> x == y' _ -> False)

mkDyn' :: Typeable a => a -> DYN mkDyn' x = DYN (x, \_ -> False)

eqDyn :: DYN -> DYN -> Bool eqDyn x@(DYN (_, fx)) y@(DYN (_,fy)) = fx y || fy x

Maybe there's some way to get mkDyn' and mkDyn as the same function, without having to re-write all of the Eq instances as a 2-parameter class like http://www.haskell.org/haskellwiki/GHC/AdvancedOverlap.

-- Adam

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-- Alberto.