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Using "deriving via" with a type family

Time:01-02

I have a typeclass with a default implementation, and would like to provide a simple way to derive the typeclass if a user wants to use their custom monad.

Here's a solution someone else provided me:

{-# 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!"

If my typeclass uses a type family, I'm unable to use deriving via. For example:

{-# 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

The last line doesn't compile:

[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’

How do I get the deriving via clause to compile when the typeclass has a type family?

CodePudding user response:

As the error message says, your associated type FooCtx depends only on ctx, but not on m, thus creating a potential for ambiguity like this:

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

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

Now it's ambiguous whether FooCtx X evaluates to Int or to String.

To fix that, just add m to parameters of 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) = ()
  ...

(I figured I'd add this as an answer since it turned out to be that simple after all)

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