Many thanks for the explanation. But I thought that GHC always derives the most generic type, why does it fix my 'a' to 'Int' ? I have another question, now that I know how to pass a generic continuation to evalAST I thought that I could use it to evaluate a more complex language: {-# LANGUAGE GADTs, RankNTypes #-} test = evalAST pr (AppA (SymA "reverse") (TxtA "abc")) t1 = eval $ App (Lit reverse) (Lit "jij") pr :: (Show a) => Expr a -> IO () pr = print . eval evalAST :: (forall a. Show a => Expr a -> IO()) -> AST -> IO () evalAST k (IntA i) = k $ Lit i evalAST k (TxtA i) = k $ Lit i evalAST k (AppA f a) = evalASTFun (\ef -> (evalAST (\ea -> k $ App ef ea) a)) f evalASTFun :: (forall a b. Expr (a-> b) -> IO ()) -> AST -> IO () evalASTFun k (SymA "reverse") = k $ Lit reverse evalASTFun k (SymA "+") = k $ Lit (+) -- An untyped expression data AST = IntA Int | TxtA String | SymA String | AppA AST AST -- A typed expression. data Expr a where Lit :: a -> Expr a App :: Expr (a->b) -> Expr a -> Expr b instance Show (a->b) where show f = "function" eval :: Expr a -> a eval (Lit i) = i eval (App f a) = (eval f) (eval a) But, this won't type check: evalAST k (AppA f a) = evalASTFun (\ef -> (evalAST (\ea -> k $ App ef ea) a)) f My understanding is that GHC correctly complains that the Expr a returned by ea is not necessarily the same Expr 'a' that ef needs. Is there any way out? Thanks, titto On 15 July 2010 11:20, Bas van Dijk <v.dijk.bas@gmail.com> wrote:
GHC tries to infer the following type for evalAST2:
evalAST2 :: forall a. (Expr a -> IO()) -> AST -> IO ()
However when the type of 'a' has been found in the first alternatives:
evalAST2 k (IntA i) = k $ Lit i
it is fixed to Int. Then the 'a' doesn't match the type (String) found in the other alternative:
evalAST2 k (TxtA i) = k $ Lit i
The reason why evalAST type checks is that the type of 'k' is
k :: forall a. (Show a) => Expr a -> IO ()
So it works not just for one 'a' but for all of them.
The way to correctly generalize evalAST is by telling GHC that 'k' indeed works for all 'a':
{-# LANGUAGE RankNTypes #-}
evalAST2 :: (forall a. Expr a -> IO()) -> AST -> IO ()
Regards,
Bas
On Thu, Jul 15, 2010 at 11:50 AM, Pasqualino "Titto" Assini <tittoassini@gmail.com> wrote:
Hi,
can anyone please explain why in the following code evalAST compiles while evalAST2 doesn't?:
Is that because the polymorphic function k is specialised in two different ways in evalAST while in evalAST2 it is constrained to be the same function?
{-# LANGUAGE GADTs #-}
test = evalAST (TxtA "abc")
-- This is OK evalAST :: AST -> IO () evalAST (IntA i) = k $ Lit i evalAST (TxtA i) = k $ Lit i
k :: (Show a) => Expr a -> IO () k e = print $ eval e
-- This is the same thing, only the k function is passed as a parameter. -- But it won't compile. -- I would expect its type to be: -- evalAST2 :: (Expr a -> IO()) -> AST -> IO () -- But is actually: -- evalAST2 :: (Expr Int -> IO ()) -> AST -> IO () evalAST2 k (IntA i) = k $ Lit i -- evalAST2 k (TxtA i) = k $ Lit i
-- The untyped expression data AST = IntA Int | TxtA String
-- A typed expression. data Expr a where Lit :: a -> Expr a
eval :: Expr a -> a eval (Lit i) = i
Thanks
titto _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe