Question

{-# LANGUAGE ExistentialQuantification, DeriveDataTypeable #-}
import Data.Typeable;

data EnumBox = forall s. (Enum s, Show s) => EB s
           deriving Typeable

instance Show EnumBox where
  show (EB s) = "EB " ++ show s

这很有效。但是，如果我想为EnumBox添加类枚举实例，请执行以下操作：

instance Enum EnumBox where
  succ (EB s) = succ s

它失败并显示以下消息：

Could not deduce (s ~ EnumBox)
from the context (Enum s, Show s)
  bound by a pattern with constructor
             EB :: forall s. (Enum s, Show s) => s -> EnumBox,
           in an equation for `succ'
  at typeclass.hs:11:9-12
  `s' is a rigid type variable bound by
      a pattern with constructor
        EB :: forall s. (Enum s, Show s) => s -> EnumBox,
      in an equation for `succ'
      at typeclass.hs:11:9
In the first argument of `succ', namely `s'
In the expression: succ s
In an equation for `succ': succ (EB s) = succ s

为什么第一个节目可以推断，但第二个节目不能？

Answer 1

你唯一的问题是succ的类型是

succ :: Enum a => a -> a

所以你需要

succ (EB s) = EB . succ $ s

再次装箱。

你也可能想要

instance Enum EnumBox where
    toEnum = EB
    fromEnum (EB i) = fromEnum i

因为这是完整性的最低定义，因为

succ = toEnum . succ . fromEnum

存在量化类型无法在类型类上下文中推断

1 个答案: