Haskell:约束中的非类型变量参数?

Question

这是代码，抱歉有点不简单。基本上，如果输入两个输入 v0 和 vs，函数 vecF 应该有两个结果 v1，v2，但由于某种原因，我最终会收到错误消息。

newtype Vector2 a = Vector2 (a,a)
deriving (Show,Eq)
newtype Vector3 a = Vector3 (a,a,a)
deriving (Show,Eq)
newtype Vector4 a = Vector4 (a,a,a,a)
deriving (Show,Eq)
class VectorSpace v where
vecZero       :: (Num a) => v a
vecSum        :: (Num a) => v a -> v a -> v a
vecScalarProd :: (Num a) => a -> v a -> v a
vecMagnitude  :: (Floating a) => v a -> a
vecInnerProd  :: (Num a) => v a -> v a-> a
instance VectorSpace Vector2 where
vecZero = Vector2 (0,0)
vecSum (Vector2 (q,w)) (Vector2 (a,s)) = Vector2 (q+a,w+s)
vecScalarProd x (Vector2 (q,w)) = Vector2 (q*x,w*x)
vecMagnitude (Vector2 (q,w)) = sqrt(q*q+w*w)
vecInnerProd (Vector2 (q,w)) (Vector2 (a,s)) = q*a+w*s
instance VectorSpace Vector3 where
vecZero = Vector3 (0,0,0)
vecSum (Vector3 (q,w,e)) (Vector3 (a,s,d)) = Vector3 (q+a,w+s,e+d)
vecScalarProd x (Vector3 (q,w,e)) = Vector3 (q*x,w*x,e*x)
vecMagnitude (Vector3 (q,w,e)) = sqrt(q*q+w*w+e*e)
vecInnerProd (Vector3 (q,w,e)) (Vector3 (a,s,d)) = q*a+w*s+e*d
instance VectorSpace Vector4 where
vecZero = Vector4 (0,0,0,0)
vecSum (Vector4 (q,w,e,r)) (Vector4 (a,s,d,f)) = Vector4 
(q+a,w+s,e+d,r+f)
vecScalarProd x (Vector4 (q,w,e,r)) = Vector4 (q*x,w*x,e*x,r*x)
vecMagnitude (Vector4 (q,w,e,r)) = sqrt(q*q+w*w+e*e+r*r)
vecInnerProd (Vector4 (q,w,e,r))(Vector4 (a,s,d,f)) =
q*a+w*s+e*d+r*f
vecF :: (Floating a, Ord a, VectorSpace v) => v a -> [v a] -> (v a, v a)
vecF v0 vs = (v1,v2) where
v0Neg     = vecScalarProd (-1) v0
v1MinMag  = minimumBy (comparing snd)(zip [0..]  
[vecMagnitude (vecSum v0Neg v) | v <- vs])
v1Index   = fst v1MinMag
v1Min     = [vecSum v0Neg v | v <- vs] !! v1Index
v1        = vecSum v0 v1Min
v2MaxMag  = maximumBy (comparing snd)(zip [0..]  
[vecMagnitude (vecSum v0Neg v) | v <- vs])
v2Index   = fst v2MaxMag
v2Max     = [vecSum v0Neg v | v <- vs] !! v2Index
v2        = vecSum v0 v2Max

以下是我输入时遇到的错误：

vecF (1,2,3,3) [(2,1,2,2),(13,3,2,1)]

.

•Non type-variable argument
in the constraint: VectorSpace ((,,,) t2 t1 t)
(Use FlexibleContexts to permit this)
•When checking the inferred type
it :: forall a t t1 t2.
(Num t, Num t1, Num t2, VectorSpace ((,,,) t2 t1 t), Ord a,
Floating a) =>
((t2, t1, t, a), (t2, t1, t, a))

我尝试过使用这样的灵活上下文，

:set -XFlexibleContexts

然后我收到这条新的错误消息

• When checking the inferred type
it :: forall a t t1 t2.
(Num t, Num t1, Num t2, VectorSpace ((,,,) t2 t1 t), Ord a,
Floating a) =>
((t2, t1, t, a), (t2, t1, t, a))
• No instance for (VectorSpace ((,,,) t2 t1 t0))
arising from a use of ‘it’
• In a stmt of an interactive GHCi command: print it

有什么方法可以解决这个问题吗？

Answer 1

您需要使用Vector4构造函数，即。

vecF (Vector4 (1,2,3,3)) [ Vector4 (2,1,2,2), Vector4 (13,3,2,1)]

基本上，错误消息准确地告诉您需要知道的内容 - 即(1,2,3,3)没有VectorSpace实例。在这种情况下，这是因为您正在调用它，而该东西不是其底层类型的向量。

它说没有VectorSpace ((,,,) t2 t1 t)实例的原因是它试图将(1, 2, 3, 4)的类型与某些a的v a相匹配 - 在这里它试图使用Num t, Num t1, Num t2 => (t, t1, t2,)作为高级位。