無法推斷（Semigroup（可選a））由實例聲明的超類引起

Question

來自“Haskell編程：來自第一原理”的以下代碼無法編譯：

module Learn where
import Data.Semigroup
import Data.Monoid

-- Exercise: Optional Monoid
data Optional a = Nada
                | Only a
                deriving (Eq, Show)

instance Monoid a => Monoid (Optional a) where
 mempty = Nada
 mappend Nada Nada = Nada
 mappend (Only a) Nada = Only $ mappend a mempty
 mappend Nada (Only a) = Only $ mappend mempty a
 mappend (Only a) (Only b) = Only $ mappend a b

它給出以下錯誤：

intermission.hs:11:10: error:
    • Could not deduce (Semigroup (Optional a))
        arising from the superclasses of an instance declaration
      from the context: Monoid a
        bound by the instance declaration at intermission.hs:11:10-40
    • In the instance declaration for ‘Monoid (Optional a)’
   |
11 | instance Monoid a => Monoid (Optional a) where
   |   

為了阻止ghc抱怨，我必須創建一個可選a的半群實例並定義“<>”。 這對我來說沒有多大意義，並且想知道是否有一些我忽略的東西。

Answer 1

“ 注意：Semigroup是Monoid的超類，因為基數為4.11.0.0。 ”

超級班的名單一直在緩慢發展。 隨着新的有用類的提出，更新舊類的API以反映它們之間的關系。 這有破壞舊代碼的不幸影響。 基礎4.11.1.0於2018年4月發布，對Monoid進行了這一重大改變。

Answer 2

這是解決方案：

import Data.Monoid

data Optional a = Nada | Only a deriving (Eq, Show)

instance Monoid a => Monoid (Optional a) where
   mempty = Nada

instance Semigroup a => Semigroup (Optional a) where
  Nada <> (Only a) = Only a
  (Only a) <> Nada = Only a
  (Only a) <> (Only a') = Only (a <> a')
  Nada <> Nada = Nada

main :: IO ()
main = do
  print $ Only (Sum 1) `mappend` Only (Sum 1)