"No instance for (Comparable Integer)" 是什么意思？

Question

data Set a = Set [a]
i1 = Set [1, 2, 3]
i2 = Set [3, 2, 1]

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 Eq 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"

执行时：cmp i1 i2我得到以下错误：

No instance for (Comparable Integer) arising from a use of ‘cmp’  
 In the expression: cmp i1 i2  
 In an equation for ‘it’: it = cmp i1 i2

我想知道这是什么意思？谢谢

Answer 1

当您执行时：

cmp i1 i2

Haskell 推断 i1 和 i2 的类型为 Set [Integer]。由于 cmp 具有类型：

cmp :: (Comparable a) => a -> a -> String

Integer 必须有一个 Comparable 类型的实例 class；它没有。因此，要解决此问题，您应该为 Integer.

提供 Comparable 的实例

另请注意，您可能打算使用 Ord 来定义不同类型值之间的比较。

"No instance for (Comparable Integer)" 是什么意思？

What does "No instance for (Comparable Integer)" mean?

haskell

typeclass