为什么我得到一个无法推断(Ord a)错误？

Question

我试图找到具有最小元素总和的列表:

shortest :: (Num a) => [[a]] -> [a]
shortest [] = []
shortest (x:xs) = if sum x < sum (shortest xs) then x else shortest xs

这给了我以下错误:

Could not deduce (Ord a) arising from a use of `<'
from the context (Eq a)
  bound by the type signature for shortest :: Eq a => [[a]] -> [a]
  at code.hs:(8,1)-(9,71)
Possible fix:
  add (Ord a) to the context of
    the type signature for shortest :: Eq a => [[a]] -> [a]
In the expression: sum x < sum (shortest xs)
In the expression:
  if sum x < sum (shortest xs) then x else shortest xs
In an equation for `shortest':
    shortest (x : xs)
      = if sum x < sum (shortest xs) then x else shortest xs

为什么没有功能类型检查？

Answer 1

此代码涉及两个类型类:Num和Ord.请注意,类型可以是成员Num而不是Ord,反之亦然.

类型sum是Num a => [a] -> a输入元素shortest需要成为的成员Num.您还在代码中执行以下操作:

sum x < sum (shortest xs)

这意味着您<在as 上使用运算符,但在您的类型签名中,您并不要求as是其实例,Ord它定义<:

class Eq a => Ord a where
  compare :: a -> a -> Ordering
  (<) :: a -> a -> Bool
  ...

因此,您需要将该要求添加到您的类型签名:

shortest :: (Ord a, Num a) => [[a]] -> [a]

或者你可以省略类型签名.

Answer 2

Num不包括Ord,所以你在类型签名中缺少Ord约束a.它应该是

shortest :: (Num a, Ord a) => [[a]] -> [a]

您可以删除类型签名,GHC将为您推断.