有没有办法将Maybe构造函数应用于具有泛型的每个记录字段？

Question

我有两种数据类型,第二种是第一种副本,但每个字段都有Maybe.

data A = {a :: Int, b :: String}
data B = {c :: Maybe Int, d :: Maybe String}

有没有办法发挥作用

f :: A -> B
g :: B -> A -> A

对领域本身没有任何了解？(如果第一个参数的值为nothing,g则将从第二个参数中取默认值)

Answer 1

这可以通过generics-sop来完成,这是一个扩展GHC默认泛型机制的库.

"generics-sop"可以定期记录并推断出它的通用表示.此表示形式具有包装每个字段的类型参数,并且库允许跨记录字段的类似应用序列的操作.

{-# language TypeOperators #-}
{-# language DeriveGeneric #-}
{-# language TypeFamilies #-}
{-# language DataKinds #-}

import qualified GHC.Generics as GHC
import Generics.SOP

data A = A {a :: Int, b :: String} deriving (Show,GHC.Generic)

instance Generic A -- this Generic is from generics-sop

defaulty :: (Generic a, Code a ~ '[ xs ]) => NP Maybe xs -> a -> a 
defaulty maybes r = case (from r) of
    SOP (Z np) -> let result = hliftA2 (\m i -> maybe i I m) maybes np
                  in  to (SOP (Z result))

main :: IO ()
main = do
   print $ defaulty (Nothing :* Just "bar" :* Nil) (A 99 "foo") 

Nothing :* Just "bar" :* Nil是与原始记录定义中的字段列表匹配的通用表示形式.请注意,表示中的每个字段都包含在中Maybe.

请参阅此处以获取generics-sop的另一个示例.