This type declaration for 'm' probably doesn't mean what you think it does. I think what you want is "m takes an item of type 'a' and returns an item of a particular type in the Num class, but I'm not going to tell you which one", but what this declaration really means "m takes an item of type 'a' and will return an item of any type you wish, so long as it is in the Num class".

In this context 'a' and 'b' are very different kinds of type variables: 'a' is fixed, but 'b' is universally quantified.

Thank you Mr. Dockins for pointing that out, now I understand (I think) why my attempts failed.

[ ... ] {-# OPTIONS -fglasgow-exts #-} class Num b => Rect a b | a -> b where width :: a -> b height :: a -> b

instance Rect IRect Int where width ( IRect (x1, _ ) (x2, _ ) ) = abs(x2 - x1) height ( IRect ( _, y1) ( _, y2) ) = abs(y2 - y1)

This works perfectly! Anyone: However, I haven't been able to make PRect an instance of this class (with extensions). I might not have grasped this yet, but I came to think; if the old class declaration would say that "width" and "height" returns something with unfixed type in the "Num" class, then wouldn't it be possible to make PRect an instance of that class (since PRect has a type parameter)? Something like this: class Rect a where width :: (Num b) => a -> b height :: (Num b) => a -> b data Num a => PRect a = PRect (a, a) (a, a) deriving (Eq, Show) instance Rect PRect a where ... This (as well as my other attemps) fail with a "Kind error: `PRect' is not applied to enough type arguments" - error. Is there a way to do it, or am I lost here? regards /johan

Am Donnerstag, 30. Juni 2005 14:07 schrieb Johan Holmquist:

[...]

Anyone:

However, I haven't been able to make PRect an instance of this class (with extensions).

If I understand your problem correctly, you may use the new Rect class (the one which is declared as class Num b => Rect a b | a -> b where ...) and add an instance Num a => Rect (PRect a) a where ...

I might not have grasped this yet, but I came to think; if the old class declaration would say that "width" and "height" returns something with unfixed type in the "Num" class, then wouldn't it be possible to make PRect an instance of that class (since PRect has a type parameter)?

Something like this:

class Rect a where width :: (Num b) => a -> b height :: (Num b) => a -> b

This means that for every type a which is an instance of Rect there is a width and a height of an arbitrary Num type. a and b are independent of each other.

data Num a => PRect a = PRect (a, a) (a, a) deriving (Eq, Show)

instance Rect PRect a where ...

This (as well as my other attemps) fail with a "Kind error: `PRect' is not applied to enough type arguments" - error. Is there a way to do it, or am I lost here?

PRect is of kind * -> *. A type has kind * if an expression can have this type. Examples of kind * types are Int, [Int], PRect Int and PRect a. Kind * -> * means that applying the type to a type of kind * yields a type of kind *. PRect (without any arguments) is an example of a kind * -> * type as well as []. From the type declarations of width an height it is clear that a has to have kind *. So you cannot use PRect for a. You could use PRect c for a but that would mean that width and height would have type Num b => PRect c -> b each, i.e. that c and b would be independent. Another solution would be the following: class Rect r where height :: Num b => r b -> b width :: Num b => r b -> b Because r is applied to b in the type of height and width, r is of kind * -> *. Now you could write instance Rect PRect where ... But you wouldn't be able to do something like instance Rect IRect where ... anymore since IRect is clearly of kind *. With this solution, you would have to have types r of kind * -> * and for every Num instance b, values of type r b would have to have a height and a width.

regards /johan

Best wishes, Wolfgang