I'm trying to create a typed expression parser in Haskell, which works great so far, but I'm currently struggling to implement higher order functions. I've boiled the problem down to a simple example:
{-# LANGUAGE TypeFamilies,GADTs,FlexibleContexts,RankNTypes #-}
-- A function has an argument type and a result type
class Fun f where
type FunArg f
type FunRes f
-- Expressions are either constants of function applications
data Expr a where
Const :: a -> Expr a
App :: Fun f => f -> FunArg f -> Expr (FunRes f)
-- A very simple function
data Plus = Plus
-- Which takes two integer expressions and returns an integer expression
instance Fun Plus where
type FunArg Plus = (Expr Int,Expr Int)
type FunRes Plus = Int
-- A more complicated function which lifts a function to lists (like in haskell)
data Map f r = Map f
-- For this we need the concept of lifting function arguments:
class Liftable a where
type LiftRes a
-- A singleton argument is lifted by changing the expression type from a to [a]
instance Liftable (Expr a) where
type LiftRes (Expr a) = Expr [a]
-- Two function arguments are lifted by lifting each argument
instance (Liftable a,Liftable b) => Liftable (a,b) where
type LiftRes (a,b) = (LiftRes a,LiftRes b)
-- Now we can declare a function instance for Map
instance (Fun f,Liftable (FunArg f),r ~ LiftRes (FunArg f)) => Fun (Map f r) where
type FunArg (Map f r) = r
type FunRes (Map f r) = [FunRes f]
-- Now a parser for functions:
parseFun :: [String] -> (forall f. Fun f => f -> a) -> a
-- The parser for the plus function is easy:
parseFun ["plus"] f = f Plus
-- But the parser for map is not possible:
parseFun ("map":sym) f
= parseFun sym (\fun -> f (Map fun))
The problem seems to be that there is no way to convince the type checker that every LiftRes is itself Liftable, because recursive class declarations are forbidden.
My question is: How do I make this work? Are there other examples of typed expression parsers from which I could take hints?
EDIT: It seems that this discussion about type family constraints seems to be very related. However, I fail to make their solution work in my case, maybe someone can help with that?