Question

我试图找到元素总和最小的列表。：

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。注意类型可以是成员Num而不是Ord，反之亦然。

sum的类型为Num a => [a] -> a，因此shortest的输入元素必须是Num的成员。您还可以执行以下操作在你的代码中：

sum x < sum (shortest xs)

这意味着您在<上使用了运算符a，但在您的类型签名中，您并不要求a是Ord的实例它定义了<：

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中：

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

或者您可以省略类型签名。

Answer 2

Num不包含Ord，因此您在类型签名中缺少Ord a约束。它应该是

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

您可以删除类型签名，GHC会为您推断。

为什么我得到一个无法推断（Ord a）错误？

2 个答案: