Hi, I was studying this post ( http://www.haskellforall.com/2012/12/the-continuation-monad.html) on CPS and I tried the following code: module Main where newtype Cont r a = Cont { runCont :: (a -> r) -> r } onInput :: Cont (IO ()) String onInput f = do s <- getLine f s onInput f main :: IO () main = onInput print I fails to compile: "Couldn't match expected type ‘Cont (IO ()) String’ with actual type ‘(String -> IO a0) -> IO b0’ • The equation(s) for ‘onInput’ have one argument, but its type ‘Cont (IO ()) String’ has none" But I thought Cont a b would be expanded to (b -> a) -> a so that Cont (IO ()) String became (String -> IO ()) -> IO (), and if I give that type using `type` instead of `newtype`, it does type-check: type Cont r a = (a -> r) -> r What am I missing here about Haskell? thanks folks!