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
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