使用类型约束时无法推断错误

Question

我正在使用Haskell实现一个线性代数示例。但是，我在声明magnitude函数时遇到了问题。

我的执行情况如下：

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

其思想是magnitude将接受Vec2D、Vec3D或Vec4D，并返回其组件的平方和的平方根。

三种向量类型中的每一种都实现了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

然而，我收到了许多错误：

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还不完全满意--我认为这是错误的间接原因。

有人能向我解释一下错误信息指向什么吗？

Answer 1

您应该使用(.)而不是($)将您的函数组合成一个管道。发生此错误的原因是，例如，Data.Foldable.foldr1 (+)希望应用于像[a]这样的Foldable类型，但实际上是将它直接应用于作为函数的fmap (^2)。

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。