Haskell 中的非法实例声明

Question

我尝试了以下代码

class Group a where
  (.+.) :: a -> a -> a
  (.-.) :: a -> a -> a
  zero :: a
  opposite :: a -> a

  x .-. y = x .+. opposite y
  opposite x = zero .-. x
  {-# MINIMAL (.+.), zero, (opposite | (.-.)) #-}

instance Fractional a => Group a where
  x .+. y = x + y
  zero = 0 :: a
  opposite = negate :: a -> a

但是在加载到 GHCi 时，出现以下错误：

group1.hs:11:26: error:
    • Illegal instance declaration for ‘Group a’
        (All instance types must be of the form (T a1 ... an)
         where a1 ... an are *distinct type variables*,
         and each type variable appears at most once in the instance head.
         Use FlexibleInstances if you want to disable this.)
    • In the instance declaration for ‘Group a’
   |
11 | instance Fractional a => Group a where
   |

我做错了什么？

Answer 1

啊！我终于明白了，哪里不对。在 Haskell 中，只能为 ADT 实例化一个 class。因此，唯一合理的解决方案是声明如下内容：

class Group a where
  (.+.) :: a -> a -> a
  (.-.) :: a -> a -> a
  zero :: a
  opposite :: a -> a

  x .-. y = x .+. opposite y
  opposite x = zero .-. x
  {-# MINIMAL (.+.), zero, (opposite | (.-.)) #-}

newtype GroupType a = GroupType a  

instance Fractional a => Group (GroupType a) where
  GroupType x .+. GroupType y = GroupType $ x + y
  zero = GroupType 0
  opposite (GroupType x) = GroupType $ negate x

Answer 2

我能够编译你的例子：

{-# LANGUAGE FlexibleInstances, UndecidableInstances #-}

class Group a where
  (.+.) :: a -> a -> a
  (.-.) :: a -> a -> a
  zero :: a
  opposite :: a -> a

  x .-. y = x .+. opposite y
  opposite x = zero .-. x
  {-# MINIMAL (.+.), zero, (opposite | (.-.)) #-}

-- data Fractional a = Fractional a a

instance (Fractional a, Num a) => Group a where
  x .+. y = x + y
  zero = 0
  opposite = negate

FlexibleInstances 允许具有约束的未知类型的实例。基本上允许instance X a
UndecidableInstances 我们需要，因为我们声明任何 a 属于 class Group 并且它可能（不可避免？）导致 a 属于Group 通过几个不同的 instance 声明。

Haskell 中的非法实例声明

Illegal instance declaration in Haskell

haskell

compiler-errors

typeclass