Re: Type Level "Application" Operator

Edward, I had the exact same thought but I couldn't get it to work. Oddly enough, I actually copied and pasted that example from my code which builds and runs perfectly well. It seems that monad transformers work differently when defined as a type (instead of newtype, which would require me to include the type parameter)? Somewhat related is the question of how to actually *export* this type alias from a module.

module Money (($)) where

type f $ x = f x infixr 0 $

doesn't work because it tries to export Prelude.$. The only way around it is to import Prelude hiding (($)). But this makes me wonder, is it actually *impossible* in Haskell to export from the same module a function with the same name at both the value and type level? Is it possible to export only one of the two? Elliot On Wed, Nov 2, 2016 at 10:42 AM, Oleg Grenrus wrote:

To make it clear:

type level `.` won’t work as an type synonym, as it’s application isn’t saturated.

{-# LANGUAGE TypeOperators #-} type (:.:) f g x = f (g x) infixr 9 :.:

type App = Maybe :.: []

fails to compile with following errors (for a reason):

• The type synonym ‘:.:’ should have 3 arguments, but has been given 2 • In the type synonym declaration for ‘App’

...

On 02 Nov 2016, at 16:24, Edward Kmett wrote:

+1, but the operator you're looking for in App there would actually be a type level version of (.).

type App a = ExceptT Err $ ReaderT Config $ LogT Text $ IO a

type App = ExceptT Err . ReaderT Config . LogT Text . IO

which would need

type (.) f g x = f (g x) infixr 9 .

to parse

-Edward

On Tue, Nov 1, 2016 at 7:13 PM, Elliot Cameron wrote: Folks,

Has there been a discussion about adding a type-level operator "$" that just mimics "$" at the value level?

type f $ x = f x infixr 0 $

Things like monad transformer stacks would look more "stack-like" with this:

type App = ExceptT Err $ ReaderT Config $ LogT Text IO

Elliot Cameron

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