Re: [Haskell-cafe] New type of ($) operator in GHC 8.0 is problematic

On Sat, Feb 06, 2016 at 01:27:00PM +0100, Ben Gamari wrote:

Tom Ellis writes:

...

On Fri, Feb 05, 2016 at 07:19:25PM +0000, Tom Ellis wrote:

...

On Fri, Feb 05, 2016 at 01:13:23PM -0500, Richard Eisenberg wrote:

...

We're in a bit of a bind in all this. We really need the fancy type for ($) so that it can be used in all situations where it is used currently.

Is there a list of situations where ($) is used currently that give rise to this need?

Does anyone have any idea about this? What is it about ($) that means it needs a new funky type whereas (apparently) nothing else does?

The first (albeit rather unconvincing) example I can think of is be something like,

getI# :: Int -> Int# getI# (I# n#) = n#

n# :: Int# n# = getI# $ 5 + 8

Richard likely has something a bit less contrived though.

I hope there's something less contrived, because if the benefit is "you get to use $ to apply functions whose return type is not of kind *" then the power to weight ratio of this is extremely low. Is it also something to do with the special treatment that $ gets in the compiler, to allow 'runST $ do'? https://www.mail-archive.com/glasgow-haskell-users@haskell.org/msg18923.html

This does raise the question of why ($) is generalized, yet (.) is not,

(.) :: forall (l :: Levity) a b (c :: TYPE l). (b -> c) -> (a -> b) -> (a -> c) (.) f g x = f (g x)

Quite.