Rebind do notation with typeclass-free monad

Question

It is possible to rebind the (>>=) and return for a monad using explicit dictionary passing like this:

{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE RebindableSyntax #-}

module Lib where

import Prelude hiding ((>>=), return)

data MonadDict m = MonadDict {
  bind :: forall a b. m a -> (a -> m b) -> m b ,
  ret :: forall a. a -> m a }

(>>=) :: (MonadDict m -> m a) -> (a -> (MonadDict m -> m b)) -> (MonadDict m -> m b)
return :: a -> (MonadDict m -> m a)

monadDictIO :: MonadDict IO

usage = let
  monadicCode = do
    ln <- const getLine 
    const . putStrLn $ ln
  in monadicCode monadDictIO

Is there a better way, how to represent the monad so one can avoid ignoring the MonadDict monad instance argument (using const) in every usage of the monadic action?

Answer 1

You can do like this:

{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE RebindableSyntax #-}
{-# LANGUAGE RecordWildCards #-}

module Lib where
import Prelude hiding(return, fail, (>>=), (>>))

data MonadDict m = MonadDict
    { (>>=)  :: forall a b. m a -> (a -> m b) -> m b
    , (>>)   :: forall a b. m a -> m b -> m b
    , return :: forall a. a -> m a
    , fail   :: forall a. String -> m a
    }

monadDictIO :: MonadDict IO
monadDictIO = ...

foo :: MonadDict m -> String -> m ()
foo = ...

usage = let
    monadicCode m@MonadDict{..} = do
        ln <- getLine
        putStrLn ln
        foo m ln
    in monadicCode monadDictIO