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!