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
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