Type Level "Application" Operator

[Ken Bateman]

Wouldn't there also be a problem with type unification?  When unifying ((f
. g) a) and (h b) do  you set ((f . g) ~ h) or ((g a) ~ b)?

On Nov 2, 2016 6:28 PM, "Edward Kmett" <ekmett at gmail.com> wrote:

> On Wed, Nov 2, 2016 at 3:11 PM, Index Int <vlad.z.4096 at gmail.com> wrote:
>
>> Edward, I don't quite follow why you think that (.) is needed here.
>> Monad transformers take two parameters, so your example is not
>> type-correct, whereas the original one is.
>>
>
> Indeed, I appear to have hyper-corrected that example.
>
> -Edward
>
> On Wed, Nov 2, 2016 at 5:24 PM, Edward Kmett <ekmett at gmail.com> 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 <eacameron at gmail.com>
>> 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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>> >>
>> >
>> >
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>
>
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