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

[Tom Ellis]

On Sat, Feb 06, 2016 at 01:27:00PM +0100, Ben Gamari wrote:
> Tom Ellis <tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk> 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.