haskell - Haskell 中的 Monoid Bool

Question

当然，数据类型并不准确，但这是否(或多或少)Monoid Bool实现了吗？

import Data.Monoid

data Bool' = T | F deriving (Show)

instance Monoid (Bool') where
    mempty = T
    mappend T _ = T
    mappend _ T = T
    mappend _ _ = F 

如果是/不是，制作 Bool 的原因是什么？的mappend一个 OR与 AND 相比?

Answer 1

有两种可能 Monoid Bool 的实例, 所以 Data.Monoid有 newtypes 来区分我们打算使用哪一个:

-- | Boolean monoid under conjunction.
newtype All = All { getAll :: Bool }
        deriving (Eq, Ord, Read, Show, Bounded, Generic)

instance Monoid All where
        mempty = All True
        All x `mappend` All y = All (x && y)

-- | Boolean monoid under disjunction.
newtype Any = Any { getAny :: Bool }
        deriving (Eq, Ord, Read, Show, Bounded, Generic)

instance Monoid Any where
        mempty = Any False
        Any x `mappend` Any y = Any (x || y)

编辑:正如Ørjan所说，实际上有四个有效实例