+1! Readability improvements are wonderful. On Tue, Nov 1, 2016 at 11:31 PM, Jon Purdy wrote:

+1 from me, if only as a nice-to-have—it would be good to flesh out some standard type operators in general.

On Nov 1, 2016 4:14 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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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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To make . work, we'd need both the ability to parse . at the type level without the compiler flipping out and assuming it is part of a rank n signature (which was the first issue I was trying to mention) and a form of "Really LiberalTypeSynonyms" like we use in ermine to allow the partial application so long as App is only used applied to an argument. In theory non recursive partial application of type synonyms within a type synonym is perfectly admissible in Haskell type checking, it just complicates the expansion and we don't do it today. Both are solvable, but are nowhere near the low hanging fruit that adding $ would be. -Edward

On 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

_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

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On Wed, Nov 2, 2016 at 3:11 PM, Index Int 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 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

_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

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Even with "ReallyLiberalTypeSynonyms" you can't have (f . g) ~ h as that would involve a partial application in a non type synonym, so there is no issue with unification. (.) only means something with all 3 arguments applied so that it can be expanded, but you can still allow it to be formally passed around inside other type synonyms so long as the final type synonym has all of its arguments expanded. type (.) f g x = f (g x) type Foo = (.) Bar Foo doesn't fully instantiate (.) but you can keep eta expanding it until it does. type Foo g x = (.) Bar g x = Bar (g x) is a perfectly legitimate definition. You can do this expansion automatically pretty easily. Given such a type synonym you can answer how many arguments it must have before it is a real type. At a use site Foo is not a type until it has been applied to two more arguments, just like the eta expanded form above. Foo ~ Baz doesn't type check for the same reason given type Id a = a you can't talk about Id ~ Bar. Id isn't a type. It needs an argument before it makes sense. This is what I mean by "ReallyLiberalTypeSynonyms". We actually wound up with these by accident in the Ermine compiler we use at work, and they turned out to be quite useful and harmless in practice. We don't have this power today, but we do have LiberalTypeSynonyms, which gets us close. -Edward On Wed, Nov 2, 2016 at 7:36 PM, Ken Bateman wrote:

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" wrote:

...

On Wed, Nov 2, 2016 at 3:11 PM, Index Int 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 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

_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

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