Type Level "Application" Operator

[Index Int]

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.

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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