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. Konrad. _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
