Bulat Ziganshin wrote:
main = do return "xx" >>= ((\x -> print x) :: Show a => a -> IO ()) main2 = do return "xx" >>= (\(x:: (forall a . (Show a) => a)) -> print x) main3 = do (x :: forall a . Show a => a) <- return "xx" print x
the second and third variant should do the same, to my mind.
Well, you need to change your mind. :-) Forall is like a function; when you write forall a. Show a => a -> IO () it means something like (a::Type) -> ShowDict a -> a -> IO () In other words, the caller supplies three things: the type a, a dictionary for Show a, and a value of a. But when you write forall a. Show a => a it means something like (a::Type) -> ShowDict a -> a In other words, the caller supplies two things, the type a and a dictionary for Show a, and the callee *returns* a value of a. In this case the callee has to be prepared to produce a value of whatever type the caller requests, and it can't do that if it only has a String. On the other hand exists a. Show a && a means (a::Type, ShowDict a, a) i.e. the callee chooses the type and dictionary as well as the value, which is what you need here. -- Ben