[Haskell-cafe] Difference between `type` and `newtype` in type checking

[Ryan Reich]

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 <roehst at gmail.com> 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!
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