如何将 Pair 定义为 Monoid？

Question

我的数据类型 Pair 定义为

data Pair a b = Pair a b

我想让它成为一个幺半群。这是我的定义。

instance (Monoid a,Monoid b) => Monoid (Pair a b) where
    mempty = Pair mempty mempty
    mappend (Pair x1 y1) (Pair x2 y2) = Pair (mappend x1 x2) (mappend y1 y2)

但是，我得到了以下错误。

foldable.hs:3:10: error:
    • Could not deduce (Semigroup (Pair a b))
        arising from the superclasses of an instance declaration
      from the context: (Monoid a, Monoid b)
        bound by the instance declaration at foldable.hs:3:10-49
    • In the instance declaration for ‘Monoid (Pair a b)’
  |
3 | instance (Monoid a,Monoid b) => Monoid (Pair a b) where
  |          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

出了什么问题？

Answer 1

mappend 不再是 Monoid 的真正方法。你可以实现它，但它应该只是 (<>) 的同义词，它在 Monoid 的超类 Semigroup 中。要实现 Monoid，您还必须实现 Semigroup，并将 mappend 的定义放在那里，命名为 (<>).

Answer 2

将来 GHC.Generics 将推出 Generically wrapper which allows deriving both instances via Generically (Pair a b)

{-# Language DeriveGeneric #-}
{-# Language DerivingStrategies #-}
{-# Language DerivingVia #-}

import GHC.Generics

-- >> mempty @(Pair String ())
-- Pair "" ()
--
-- >> Pair "hello" () <> Pair " " undefined <> Pair "world" ()
-- Pair "hello world" ()
data Pair a b
  deriving
  stock Generic

  deriving (Semigroup, Monoid)
  via Generically (Pair a b)

generic-data.

已经可以做到这一点

如何将 Pair 定义为 Monoid？

How can I define Pair as a Monoid?

haskell

monoids