Continuation monad for a yield/await function in Haskell

Question

I want to create an automata type with a type like this:

newtype Auto i o = Auto {runAuto :: i -> (o, Auto i o)}

I know this is the type

Answer 1

You could extend your automaton in the spirit of Conduit, that is, allow it to exit and return a value on finitely many inputs:

data Auto i o a
    = Step (i -> (o, Auto i o a))
    | End a

Then you can define a monad instance that concatenates two automata using >>=: When the first one finishes, the second one continues.

The good news is that you don't need to implement it yourself. Returning a value or nesting using a functor is exactly what the free monad does (see its haddock docs). So let's define

{-# LANGUAGE DeriveFunctor #-}
import Control.Monad.Free
import Data.Void

-- | A functor describing one step of the automaton
newtype AutoF i o t = AutoF (i -> (o, t))
  deriving (Functor)

Then the original Auto type can be defined just as an alias:

type Auto i o = Free (AutoF i o)

This automatically gives you all the instances of Free, and you can also define your original functions:

-- | If @a@ is 'Void', the machine runs forever:
runAuto :: Auto i o Void -> i -> (o, Auto i o Void)
runAuto (Pure v)  _         = absurd v
runAuto (Free (AutoF f)) i  = f i

yield :: o -> Auto i o ()
yield x = liftF (AutoF $ \_ -> (x, ()))

Note that using the same functor with FreeT you get the corresponding monad transformer:

import Control.Monad.Trans.Free

type AutoT i o = FreeT (AutoF i o)

yieldT :: (Monad m) => o -> AutoT i o m ()
yieldT x = liftF (AutoF $ \_ -> (x, ()))

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