Re: Type Level "Application" Operator

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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