On Thursday 10 June 2010 22:01:38, Dupont Corentin wrote:
> Hello Maciej,
> i tried this out, but it didn't worked.
> 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
> > 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 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)
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
> > 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,
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