haskell - 是否可以使用 Free 在 DSL 中实现多态函数

Answer 1

You can represent your DSL with a GADT

{-# LANGUAGE GADTs #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE DeriveFunctor #-}

data Queue a = Queue a

data MyDsl next where
  NewQueue :: (Queue a -> next) -> MyDsl next
  WriteToQueue :: (Queue a) -> a -> next -> MyDsl next

deriving instance Functor MyDsl

Neither makeFree nor makeFreeCon can generate free polymorphic monadic actions for MyDsl. You will need to write them yourself.

{-# LANGUAGE FlexibleContexts #-}

import Control.Monad.Free.Class

newQueue :: (MonadFree MyDsl m) => m (Queue a)
newQueue = wrap $ NewQueue return

writeToQueue :: (MonadFree MyDsl m) => Queue a -> a -> m ()
writeToQueue q v = liftF $ WriteToQueue q v ()

Now you can write your test program.

{-# LANGUAGE ScopedTypeVariables #-}

import Control.Monad.Free.Church

-- testProgram can have a more general type
-- testProgram :: (MonadFree MyDsl m) => m ()
testProgram :: F MyDsl ()
testProgram = do
  (intQueue :: Queue Int) <- newQueue
  (charQueue :: Queue Char) <- newQueue
  writeToQueue intQueue 1
  writeToQueue charQueue 'c'

---

You may find your DSL is easier to write multiple interpreters for if you parameterize the type of a queue. If you do, you'll need a type family or functional dependency to determine the type of the queue from the type of the monad.

data MyDsl q next where
  NewQueue :: (q a -> next) -> MyDsl next
  WriteToQueue :: (q a) -> a -> next -> MyDsl next