tInfer :: MonadTypeInfer m => Expr -> m Type eval :: MonadEval m => Expr -> m Expr
That solves!
I should've left out type annotation.
On Mon, May 12, 2008 at 10:38 AM, Twan van Laarhoven
sam lee wrote:
Hi.
I want to compose two monads to build another monad where computations of the two monads can be used inside.
I have:
- MonadTypeInfer : interface (class) for TypeInfer monad - TypeInfer : a monad that has Map String Type (association of names and
types)
- TypeInferT : transformer of above monad - MonadEval : interface (class) for Eval monad - Eval : a monad that has Map String Expr (association of names and code/function body) - EvalT : transformer of Eval - tInfer :: Expr -> TypeInfer Type -- given expr, returns type of it in TypeInfer monad - eval :: Expr -> Eval Expr -- given expr, returns normalized expr in Eval monad
Is there a way to build a monad where you could use sub-monads' (monads used to build current monad) computations?
A solution to this problem is to use type classes, and in particular MonadTrans. You can then give an instance of MonadTypeInfer for EvalT m where m is an instance of MonadTypeInfer, and similarly an instance MonadEval for TypeInferT m. How this is implemented depends on the Monads in question, but if you use the monad transformer library with newtype deriving you can just add "deriving MonadTrans".
class Monad m => MonadTypeInfer m where -- functions -- tiStuff :: X -> m Whatever
class Monad m => MonadEval m where -- functions --
instance Monad m => MonadTypeInfer (TypeInferT m) where -- implementation -- tiStuff = ...
instance Monad m => MonadEval (EvalT m) where -- implementation --
instance MonadEval m => MonadTypeInfer (EvalT m) where -- lift the functions from TypeInfer through the EvalT type, -- MonadTrans from the mtl might help here tiStuff x = lift (tiStuff x)
tInfer :: MonadTypeInfer m => Expr -> m Type eval :: MonadEval m => Expr -> m Expr
Twan