Question

这是一个与我的模块here相关的问题，并且有点简化。它也与this之前的问题有关，我在其中过度简化了我的问题并没有得到我想要的答案。我希望这不是太具体，如果你能想到更好的话，请更改标题。

背景

我的模块使用并发chan，分为读取侧和写入侧。我使用一个带有关联类型同义词的特殊类来支持多态通道“连接”：

{-# LANGUAGE TypeFamilies #-}

class Sources s where
    type Joined s
    newJoinedChan :: IO (s, Messages (Joined s)) -- NOT EXPORTED

--output and input sides of channel:
data Messages a  -- NOT EXPORTED
data Mailbox a

instance Sources (Mailbox a) where
    type Joined (Mailbox a) = a
    newJoinedChan = undefined

instance (Sources a, Sources b)=> Sources (a,b) where
    type Joined (a,b) = (Joined a, Joined b)
    newJoinedChan = undefined

-- and so on for tuples of 3,4,5...

上面的代码允许我们做这样的事情：

example = do
    (mb ,        msgsA) <- newJoinedChan
    ((mb1, mb2), msgsB) <- newJoinedChan
    --say that: msgsA, msgsB :: Messages (Int,Int)
    --and:      mb :: Mailbox (Int,Int)
    --          mb1,mb2 :: Mailbox Int

我们有一个名为Behavior的递归操作，我们可以对从频道的“读取”端拉出的消息进行操作：

newtype Behavior a = Behavior (a -> IO (Behavior a))
runBehaviorOn :: Behavior a -> Messages a -> IO ()  -- NOT EXPORTED

这样我们就可以在Behavior (Int,Int)或msgsA上运行msgsB，在第二种情况下，它收到的元组中的Int个实际上都来了单独Mailbox es。

对于暴露的spawn函数

中的用户，这一切都联系在一起

spawn :: (Sources s) => Behavior (Joined s) -> IO s

...调用newJoinedChan和runBehaviorOn，并返回输入Sources。

我想做什么

我希望用户能够创建Behavior任意产品类型（而不仅仅是元组），因此我们可以在上面的示例Behavior (Pair Int Int)上运行Messages 。我希望GHC.Generics使用Sources同时使用多态spawn :: (Sources s, Generic (Joined s), Rep (Joined s) ~ ??) => Behavior (Joined s) -> IO s，但无法使其正常工作。

fst

API中实际公开的上述示例部分是newJoinedChan操作的Behavior和runBehaviorOn s，因此可接受的解决方案可以修改其中一个或全部snd或newJoinedChan的{{1}}。

我还将上面的API扩展到支持总和（尚未实现），如Behavior (Either a b)，所以我希望GHC.Generics能为我工作。

问题

我是否有办法扩展上述API以支持任意Generic a=> Behavior a？
如果不使用GHC的Generics，还有其他方法可以获得我想要的API，同时最小化最终用户的痛苦（即他们只需要在其类型中添加派生子句）吗？例如与Data.Data？

Answer 1

也许是这样的？

{-# LANGUAGE TypeFamilies, DeriveGeneric, DefaultSignatures, TypeOperators, FlexibleInstances, FlexibleContexts, UndecidableInstances #-}

import Control.Arrow
import GHC.Generics

class Sources s where
    type Joined s
    newJoinedChan :: IO (s, Messages (Joined s)) -- NOT EXPORTED
    default newJoinedChan :: (Generic s, SourcesG (Rep s)) => IO (s, Messages (JoinedG (Rep s)))
    newJoinedChan = fmap (first to) newJoinedChanG

class SourcesG g where
    type JoinedG g
    newJoinedChanG :: IO (g a, Messages (JoinedG g))

--output and input sides of channel:
data Messages a  -- NOT EXPORTED
data Mailbox a

instance Sources (Mailbox a) where
    type Joined (Mailbox a) = a
    newJoinedChan = undefined

instance (Sources a, Sources b)=> Sources (a,b) where
    type Joined (a,b) = (Joined a, Joined b)
    newJoinedChan = undefined

instance (SourcesG a, SourcesG b) => SourcesG (a :*: b) where
    type JoinedG (a :*: b) = (JoinedG a, JoinedG b)
    newJoinedChanG = undefined

instance (SourcesG a, Datatype c) => SourcesG (M1 D c a) where
    type JoinedG (M1 D c a) = JoinedG a
    newJoinedChanG = fmap (first M1) newJoinedChanG

instance (SourcesG a, Constructor c) => SourcesG (M1 C c a) where
    type JoinedG (M1 C c a) = JoinedG a
    newJoinedChanG = fmap (first M1) newJoinedChanG

instance (SourcesG a, Selector c) => SourcesG (M1 S c a) where
    type JoinedG (M1 S c a) = JoinedG a
    newJoinedChanG = fmap (first M1) newJoinedChanG

instance Sources s => SourcesG (K1 i s) where
    type JoinedG (K1 i s) = Joined s
    newJoinedChanG = fmap (first K1) newJoinedChan

newtype Behavior a = Behavior (a -> IO (Behavior a))

runBehaviorOn :: Behavior a -> Messages a -> IO ()
runBehaviorOn = undefined

spawn :: (Sources s) => Behavior (Joined s) -> IO s
spawn = undefined

data Pair a b = Pair a b deriving (Generic)

instance (Sources a, Sources b) => Sources (Pair a b) where
    type Joined (Pair a b) = JoinedG (Rep (Pair a b))

使用GHC.Generics或Data.Data键入Families

背景

我想做什么

问题

1 个答案: