# More type design questions

**Andre Pang
**
ozone@algorithm.com.au

*Tue, 19 Aug 2003 04:04:19 +1000*

On Tuesday, August 19, 2003, at 03:33 AM, Konrad Hinsen wrote:
>* On Monday 18 August 2003 19:10, Andre Pang wrote:
*>*
*>>* This seems to work (with -fglasgow-exts):
*>>*
*>>* module Foo where
*>>*
*>>* class Vect v where
*>>* (<+>) :: v -> v -> v
*>>*
*>>* data Vector a = Vector a a a
*>>* deriving (Show, Eq)
*>>*
*>>* instance Floating a => Vect (Vector a) where
*>>* (<+>) (Vector x1 y1 z1) (Vector x2 y2 z2)
*>>* = Vector (x1+x2) (y1+y2) (z1+z2)
*>>*
*>>* instance Floating a => Vect [Vector a] where
*>>* (<+>) l1 l2 = zipWith (<+>) l1 l2
*>>*
*>>* *Foo> (Vector 5 6 7) <+> (Vector 1 2 3)
*>>* Vector 6.0 8.0 10.0
*>>* *Foo> [Vector 1 2 3, Vector 10 20 30] <+> [Vector 100 200 300, Vector
*>>* 4
*>>* 5 6]
*>>* [Vector 101.0 202.0 303.0,Vector 14.0 25.0 36.0]
*>>*
*>>* ... or does example not do something which you want it to do?
*>*
*>* Well, yes, because my original example was cut down to illustrate the
*>* problem
*>* I had. The full version of the class Vect is
*>*
*>* class Vect v a where
*>* (<+>) :: Floating a => v a -> v a -> v a
*>* (<->) :: Floating a => v a -> v a -> v a
*>* (<*>) :: Floating a => a -> v a -> v a
*>*
*>* I need the parametrization on a in order to be able to define the type
*>* of
*>* scalar multiplication.
*>*
*>* I do have the choice of "class Vect v" or "class Vect v a", both seem
*>* to do
*>* the same in this context, but in both cases "v" has the role of a type
*>* constructor.
*
Ah. What about the code I gave above, and in addition to that:
class (Floating a, Vect v) => VectMult v a where
(<*>) :: a -> v -> v
instance VectMult (Vector Float) Float where
(<*>) n (Vector x y z) = Vector (n*x) (n*y) (n*z)
?
--
% Andre Pang : trust.in.love.to.save