Although a newtype has the same representation at runtime as the type inside, you still have to use the constructor as for a 'data' type. This makes sense, as it is a *new type*, whose values must be distinguished from those of the inner type, and the only way to do that is if they are decorated with a constructor. On Mon, Sep 10, 2018 at 5:55 PM Rodrigo Stevaux wrote:

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! _______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.