Question

data Set a = Set [a]
member xs x = elem x xs
subset xs ys = and (map (member ys) xs)

instance (Eq a) => Eq (Set a) where  
  (Set xs) == (Set ys) = (subset xs ys) && (subset ys xs)

class Ord a => Comparable a where
  cmp :: a -> a -> String
  cmp x y
    | x == y    = "eq"
    | otherwise = "neq"

instance Comparable a => Comparable (Set a) where
  cmp (Set xs) (Set ys)
    | (Set xs) == (Set ys) = "same"
    | otherwise            = "different"

我收到以下错误：

无法演绎（Ord（Set a））来自实例声明的超类从上下文（可比较的a）由'Comparable（Set a）'

的实例声明绑定

我想知道错误是什么？感谢。

Answer 1

Comparable是Ord的子类，因此您需要为Ord (Set a)添加一个实例，例如：

instance Ord a => Ord (Set a) where
  compare (Set s1) (Set s2) =
    compare (sort s1) (sort s2)

Answer 2

此声明

class Ord a => Comparable a where

声明类Comparable的每个成员都必须是类Ord的成员。如果我要尝试

data MyType = ...
instance Comparable MyType where
  cmp x y = "hello"

编译器会抱怨缺少实例Ord MyType。我没有使用Ord实例中的任何内容并不重要：Comparable的定义需要它。

要解决此问题，请添加实例

instance Ord a => Ord (Set a) where
   compare setA setB = ....

之后，如果您向约束添加Comparable，则可以使用Ord a实例。

instance (Ord a , Comparable a) => Comparable (Set a) where
  cmp (Set xs) (Set ys)
    | (Set xs) == (Set ys) = "same"
    | otherwise            = "different"

Answer 3

您为设置数据类型提供了Equality实例，但它没有Ord实例。当您尝试创建Comparable类的Set和实例时，它要求Set必须是Ord的实例。

Haskell - 编译错误

3 个答案: