I come up a solution as this:
---------------------------------------------------
module Vector where
data Vector =Vector [Double]
fromVector :: Vector -> [Double]
fromVector (Vector v) = v
fromList :: [Double] -> Vector
fromList v = Vector v
toVector :: Double -> Vector
toVector x = Vector (repeat x)
instance Eq Vector where
v1 ==v2 = (fromVector v1) == (fromVector v2)
instance Show Vector where
show v = show (fromVector v)
instance Num Vector where
v1 + v2 =Vector (zipWith (+) (fromVector v1) (fromVector v2))
v1 - v2 =Vector (zipWith (-) (fromVector v1) (fromVector v2))
v1 * v2 =Vector (zipWith (*) (fromVector v1) (fromVector v2))
signum v = Vector (map signum (fromVector v))
abs v = Vector ((repeat.sqrt.sum.fromVector) (v*v))
fromInteger n =Vector (repeat (fromInteger n))
instance Fractional Vector where
v1 / v2 = Vector (zipWith (/) (fromVector v1) (fromVector v2))
fromRational r =Vector (repeat (fromRational r))
--------------------------------------------------------------------------
rk4 :: ((Vector,Vector)->Vector)->Vector->Vector->Vector->[Vector]
rk4 _ _ _ (Vector []) = []
rk4 f h y0 (Vector (x0:xs)) = y0 :rk4 f h y1 (Vector xs) where y1=yp f h (toVector
x0) y0
yp ::((Vector,Vector)->Vector)->Vector->Vector->Vector->Vector
yp f h x y = y + (k1 + 2 * (k2 + k3) + k4)
where k1=h*f(x,y)
k2=h*f(x+0.5*h, y +(0.5*k1))
k3=h*f(x+0.5*h, y +(0.5*k2))
k4=h*f(x+h, y+k3)
a=let g (x,y1) = y1
x0 = 0
h = 0.01
x =Vector [x0,x0+h..3]
y0 =Vector [0,0.5]
in rk4 g (toVector h) y0 x
--------------------------------------------------------------------------
The main problem is how to make type convert implicitly.
Whem a function needs a vector as its parameter, pass a double and it is
converted to vector implicitly.
======= 2003-07-12 12:18:00 Jon Fairbairn Wrote：=======
>On 2003-07-12 at 20:20+1000 Andrew J Bromage wrote:
>> G'day all.
>>
>> On Fri, Jul 11, 2003 at 04:28:19PM -0400, Dylan Thurston wrote:
>>
>> > Don't be silly [...]
>>
>> Never!
>
>Or only sometimes. I'm surprised that no-one has yet
>answered the question "How overload operator in Haskell?"
>with "Overload operator in Haskell fine". (cf Cary Grant)
I am also surprised at this, it can be done by C++ .
>
>--
>J? Fairbairn [EMAIL PROTECTED]
>31 Chalmers Road [EMAIL PROTECTED]
>Cambridge CB1 3SZ +44 1223 570179 (after 14:00 only, please!)
>
>
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= = = = = = = = = = = = = = = = = = = =
Regards,
Liu Junfeng
[EMAIL PROTECTED]
2003-07-12
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