Question

我正在使用Haskell来实现线性代数示例。但是，在声明magnitude函数时遇到了问题。

我的实施如下：

magnitude :: (Foldable t, Functor t, Floating a) => t a -> a
magnitude = sqrt $ Data.Foldable.foldr1 (+) $ fmap (^2)

我们的想法是magnitude会接受Vec2D，Vec3D或Vec4D，并返回其组成部分的平方和的平方根。

三种矢量类型中的每一种都实现Functor和Foldable。例如，

newtype Vec2D = Vec2D (a, a) deriving (Eq, Show)
instance Functor Vec2D where
    fmap f (Vec2D (x, y)) = Vec2D (f x, f y)
instance Foldable Vec2D where
    foldr f b (Vec2D (x, y)) = f x $ f y b

但是，我收到了很多错误：

LinearAlgebra.hs:9:13:
    Could not deduce (Floating (t a -> a)) arising from a use of `sqrt'
    from the context (Foldable t, Functor t, Floating a)
      bound by the type signature for
                 magnitude :: (Foldable t, Functor t, Floating a) => t a -> a
      at LinearAlgebra.hs:8:14-60
    Possible fix: add an instance declaration for (Floating (t a -> a))
    In the expression: sqrt
    In the expression: sqrt $ Data.Foldable.foldr1 (+) $ fmap (^ 2)
    In an equation for `magnitude':
        magnitude = sqrt $ Data.Foldable.foldr1 (+) $ fmap (^ 2)

LinearAlgebra.hs:9:20:
    Could not deduce (Foldable ((->) (t a -> a)))
      arising from a use of `Data.Foldable.foldr1'
    from the context (Foldable t, Functor t, Floating a)
      bound by the type signature for
                 magnitude :: (Foldable t, Functor t, Floating a) => t a -> a
      at LinearAlgebra.hs:8:14-60
    Possible fix:
      add an instance declaration for (Foldable ((->) (t a -> a)))
    In the expression: Data.Foldable.foldr1 (+)
    In the second argument of `($)', namely
      `Data.Foldable.foldr1 (+) $ fmap (^ 2)'
    In the expression: sqrt $ Data.Foldable.foldr1 (+) $ fmap (^ 2)

LinearAlgebra.hs:9:41:
    Could not deduce (Num (t a -> a)) arising from a use of `+'
    from the context (Foldable t, Functor t, Floating a)
      bound by the type signature for
                 magnitude :: (Foldable t, Functor t, Floating a) => t a -> a
      at LinearAlgebra.hs:8:14-60
    Possible fix: add an instance declaration for (Num (t a -> a))
    In the first argument of `Data.Foldable.foldr1', namely `(+)'
    In the expression: Data.Foldable.foldr1 (+)
    In the second argument of `($)', namely
      `Data.Foldable.foldr1 (+) $ fmap (^ 2)'
Failed, modules loaded: none.

我对Functor或Foldable感到不舒服 - 我相信这是错误的间接原因。

有人可以向我解释错误消息指向的内容吗？

Answer 1

您应该将您的功能合并到(.)而不是($)的管道中。发生此错误的原因是，例如，Data.Foldable.foldr1 (+)期望应用于Foldable类型[a]，但您实际上是直接将其应用于fmap (^2)这是一个函数

magnitude :: (Foldable t, Functor t, Floating a) => t a -> a
magnitude = sqrt . Data.Foldable.foldr1 (+) . fmap (^2)

或

magnitude :: (Foldable t, Functor t, Floating a) => t a -> a
magnitude ta = sqrt $ Data.Foldable.foldr1 (+) $ fmap (^2) $ ta

两者都会做得更好。

使用类型约束时无法推断错误

1 个答案: