將“派生通過”與型別系列一起使用

我有一個帶有默認實作的型別類，如果用戶想要使用他們的自定義 monad，我想提供一種派生型別類的簡單方法。

這是別人為我提供的解決方案：

{-# LANGUAGE DerivingVia #-}
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE StandaloneDeriving #-}

import Control.Monad.Cont (MonadIO, MonadTrans (lift))
import Control.Monad.Reader (MonadReader, ReaderT (runReaderT))

----------------- My module's definitions -----------------

class Monad m => MonadFoo m where
  foo :: m ()

instance MonadFoo IO where
  foo = putStrLn "Hello world!"

instance MonadFoo m => MonadFoo (ReaderT r m) where
  foo = lift foo

------------------------------------------------------------

------ The user's custom monad   instance definitions ------

data AppEnv = AppEnv

newtype AppM a = AppM
  { runAppM :: ReaderT AppEnv IO a
  }
  deriving (Functor, Applicative, Monad, MonadIO, MonadReader AppEnv)

deriving via (ReaderT AppEnv IO) instance MonadFoo AppM

------------------------------------------------------------

-- Example usage
program :: IO ()
program = runReaderT (runAppM foo) AppEnv

> program
"Hello world!"

如果我的型別類使用型別系列，我將無法使用通過. 例如：

{-# LANGUAGE AllowAmbiguousTypes #-}
{-# LANGUAGE DerivingVia #-}
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE InstanceSigs #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE TypeApplications #-}
{-# LANGUAGE TypeFamilies #-}

import Control.Monad.Cont (MonadIO, MonadTrans (lift))
import Control.Monad.Reader (MonadReader, ReaderT (runReaderT))

----------------- My module's definitions -----------------

class Monad m => MonadFoo ctx m where
  type FooCtx ctx
  foo :: m (FooCtx ctx)

data DummyCtx = DummyCtx

instance MonadFoo DummyCtx IO where
  type FooCtx DummyCtx = ()
  foo :: IO ()
  foo = putStrLn "hello"

instance MonadFoo DummyCtx m => MonadFoo DummyCtx (ReaderT r m) where
  type FooCtx DummyCtx = ()
  foo :: ReaderT r m ()
  foo = lift $ foo @DummyCtx

------------------------------------------------------------

------ The user's custom monad   instance definitions ------

data AppEnv = AppEnv

newtype AppM a = AppM
  { runAppM :: ReaderT AppEnv IO a
  }
  deriving (Functor, Applicative, Monad, MonadIO, MonadReader AppEnv)

deriving via (ReaderT AppEnv IO) instance MonadFoo DummyCtx AppM

最后一行不編譯：

[typecheck] [E] ? Can't make a derived instance of                                                                                       
~          ‘MonadFoo DummyCtx AppM’ with the via strategy:                                                                                      
~          the associated type ‘FooCtx’ is not parameterized over the last type                                                                 
~      variable                                                                                                                                 
~            of the class ‘MonadFoo’                                                                                                            
~      ? In the stand-alone deriving instance for ‘MonadFoo DummyCtx AppM’

當型別類具有型別系列時，如何讓派生 via子句進行編譯？

uj5u.com熱心網友回復：

正如錯誤訊息所說，您的關聯型別FooCtx僅取決于ctx，而不取決于m，因此可能會產生這樣的歧義：

instance MonadFoo X A where
  type FooCtx X = Int
  ...

instance MonadFoo X B where
  type FooCtx X = String
  ...

現在不明確是FooCtx X評估為Int還是String。

要解決這個問題，只需添加m到以下引數FooCtx：

class Monad m => MonadFoo ctx m where
  type FooCtx ctx m
  ...

instance MonadFoo DummyCtx IO where
  type FooCtx DummyCtx IO = ()
  ...

instance MonadFoo DummyCtx m => MonadFoo DummyCtx (ReaderT r m) where
  type FooCtx DummyCtx (ReaderT r m) = ()
  ...

（我想我會添加這個作為答案，因為它畢竟是那么簡單）

標籤：哈斯克尔 推导 导出通过

上一篇：haskell中是否有包含字串和串列的型別？

下一篇：Haskell嵌套函式順序