![](/img/trans.png)
[英]A basic Monoid definition gives "No instance for (Semigroup MyMonoid) arising from the superclasses of an instance declaration"
[英]Could not deduce (Semigroup (Optional a)) arising from the superclasses of an instance declaration
来自“Haskell编程:来自第一原理”的以下代码无法编译:
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
它给出以下错误:
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
|
为了阻止ghc抱怨,我必须创建一个可选a的半群实例并定义“<>”。 这对我来说没有多大意义,并且想知道是否有一些我忽略的东西。
“ 注意:Semigroup是Monoid的超类,因为基数为4.11.0.0。 ”
超级班的名单一直在缓慢发展。 随着新的有用类的提出,更新旧类的API以反映它们之间的关系。 这有破坏旧代码的不幸影响。 基础4.11.1.0于2018年4月发布,对Monoid进行了这一重大改变。
这是解决方案:
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)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.