Difference between `type` and `newtype` in type checking

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Difference between `type` and `newtype` in type checking

Rodrigo Stevaux
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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Re: 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 <[hidden email]> 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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Only members subscribed via the mailman list are allowed to post.

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Only members subscribed via the mailman list are allowed to post.