On Thursday 10 June 2010 22:01:38, Dupont Corentin wrote:
Hello Maciej, i tried this out, but it didn't worked.
Daniel,
I added a (Show a) constraint to Equal:
data Obs a where Player :: Obs Integer Turn :: Obs Integer Official :: Obs Bool Equ :: (Show a, Eq a) => Obs a -> Obs a -> Obs Bool
--woops!!
Plus :: (Num a) => Obs a -> Obs a -> Obs a Time :: (Num a) => Obs a -> Obs a -> Obs a Minus :: (Num a) => Obs a -> Obs a -> Obs a Konst :: a -> Obs a And :: Obs Bool -> Obs Bool -> Obs Bool Or :: Obs Bool -> Obs Bool -> Obs Bool
It works for the Show instance, but not Eq. By the way, shouldn't the Show constraint be on the instance and not on the datatype declaration?
Can't be here, because of Equ :: Obs a -> Obs a -> Obs Bool You forget the parameter a, and you can't recover it in the instance declaration. So you have to provide the Show instance for a on construction, i.e. put the constraint on the data constructor.
I'd prefer to keep the datatype as generic as possible...
There is really no way to make my Obs datatype an instance of Eq and Show??
Show can work (should with the constraint on Equ), Eq is hairy. instance Show t => Show (Obs t) where show (Equ a b) = show a ++ " `Equal` " ++ show b show (Plus a b) = ... show (Konst x) = "Konst " ++ show x ... For an Eq instance, you have the problem that Equ (Konst True) (Konst False) and Equ Player Turn both have the type Obs Bool, but have been constructed from different types, so you can't compare (Konst True) and Player. I don't see a nice way to work around that.
I searched around a way to add type information on the pattern match
like:
instance Show t => Show (Obs t) where show (Equal (a::Obs t) (b::Obs t)) = (show a) ++ " Equal " ++ (show b) show (Plus a b) = (show a) ++ " Plus " ++ (show b)
But it doesn't work.
thanks for your help, Corentin