Am Donnerstag 09 April 2009 00:29:24 schrieb Marco Túlio Gontijo e Silva:
Hello,
I'm getting this error message from GHC in the following code:
type M a = Monad m => m a
class A a b where f :: Monad m => a -> m b
instance A String Char where f string = return $ head string
instance A Char Int where f int = return $ fromEnum int
I thought at first to write h like this, but it gives me the error. If I write it like the other h uncommented, there's no error. I can't see why they aren't equivalent.
h :: IO () h = f "abc" >>= (f :: Char -> M Int) >>= print
h :: IO () h = (f "abc" :: M Char) >>= f >>= (print :: Int -> IO ())
Can someone clarify this to me?
Greetings.
The first error message when loading that code is Inferred.hs:3:0: Illegal polymorphic or qualified type: forall (m :: * -> *). (Monad m) => m a Perhaps you intended to use -XRankNTypes or -XRank2Types In the type synonym declaration for `M' That can be a hint. The type synonym M is probably not what you think. This: ---------------------------------------------------------------- class B a b where foo :: a -> M b instance B String Char where foo str = return $ head str instance B Char Int where foo c = return $ fromEnum c k :: IO () k = foo "abc" >>= (foo :: Char -> M Int) >>= print ---------------------------------------------------------------- works, as does m = foo "abc" >>= (foo :: Char -> (forall x. Monad x => x Int)) >>= print And that reveals what's going on, since m is in fact the same as k. The type of f in class A says that given any specific monad m and a value of type a, f can produce a value of type m b. But when you write (f :: Char -> M Int) , you say that given a Char, f produces a value which belongs to m Int *for every monad m*. Thus the type you state for f (the expected type) is more polymorphic than the actual type f has according to the class definition (which is the inferred type).