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 <novadenizen@gmail.com> 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" <ekmett@gmail.com> wrote:
On Wed, Nov 2, 2016 at 3:11 PM, Index Int <vlad.z.4096@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@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@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
_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries
_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries
_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries