如何理解这个class类型的错误告诉我我用错了类型？

Question

这是我计算一列数字的平均值的函数

arithMean :: (Fractional a) => [a] -> a
arithMean list = (foldr (+) 0 list)/ (length list)

然而，它引发了以下错误

HigherOrder.hs:10:39: error:
? Couldn't match expected type ‘a’ with actual type ‘Int’
  ‘a’ is a rigid type variable bound by
    the type signature for:
      arithMean :: forall a. Fractional a => [a] -> a
    at HigherOrder.hs:9:1-39
? In the second argument of ‘(/)’, namely ‘(length list)’
  In the expression: (foldr (+) 0 list) / (length list)
  In an equation for ‘arithMean’:
      arithMean list = (foldr (+) 0 list) / (length list)
? Relevant bindings include
    list :: [a] (bound at HigherOrder.hs:10:11)
    arithMean :: [a] -> a (bound at HigherOrder.hs:10:1)

关于这个错误，我很困惑为什么它强制我将"list"更改为[Int]，我认为无论是什么类型，它都应该有效

此外，我认为当我尝试使用 "length" 函数时可能发生了一些事情，因为当我在下面编写这段代码时，它起作用了

arithMean :: (Fractional a) => [a] -> a
arithMean list = (foldr (+) 0 list)

Answer 1

(/) :: Fractional a => a -> a -> a means that the numerator and denominator need to be both need to have the same Fractional类型的类型。

length :: [a] -> Int 但是将可折叠映射到 Int。一个 Int 是不是一个 Fractional 类型。

您可以使用 fromIntegral :: (Integral a, Num b) :: a -> b here to convert the Int to an type that is a member of the Num 类型类。

arithMean :: (Fractional a, Foldable f) => f a -> a
arithMean list = sum list / <b>fromIntegral</b> (length list)

我们可以在这里使用sum :: (Foldable f, Num a) => f a -> a代替foldr (+) 0 list。

如何理解这个class类型的错误告诉我我用错了类型？

How to understand this type class error of telling me I use the wrong type?

haskell

types

typeclass