Haskell的Monoid Bool

Question

当然数据类型并不精确,但这是(或多或少)Monoid Bool实现的方式？

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,什么是做推理Bool的mappend一OR对AND？

Answer 1

有两种可能的Monoid实例Bool,因此新Data.Monoid类型可以区分我们打算使用哪种实例:

-- | 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:实际上有四个有效的实例,如Ørjan笔记

Answer 2

您提供的实例不是幺半群。

mappend F mempty
mappend F T  -- by definition of mempty
T            -- by definition of mappend

所以我们已经证明了F <> mempty === T，但是对于任何幺半群，x <> mempty === x.

的单位Any是False，单位All是True。