Thanks you all, now it makes sense.
titto
On 15 July 2010 17:52, Brent Yorgey
On Thu, Jul 15, 2010 at 01:20:11PM +0100, Pasqualino Titto Assini wrote:
Many thanks for the explanation.
But I thought that GHC always derives the most generic type, why does it fix my 'a' to 'Int' ?
Note that this type
evalAST2 :: forall a. (Expr a -> IO()) -> AST -> IO ()
means that the type a has to be chosen first, i.e. by the *caller* of evalAST2. So evalAST2 is not free to use its argument k at whatever type it wants, it will be given a function of type (Expr T -> IO()) for some type T that it didn't get to choose. Forcing the caller to provide a polymorphic function, so that evalAST2 can choose at what type(s) to use it, is exactly what is expressed by the rank-2 type
evalAST2 :: (forall a. Expr a -> IO()) -> AST -> IO ()
-Brent _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
-- Pasqualino "Titto" Assini, Ph.D. http://quicquid.org/