[Haskell-cafe] New type of ($) operator in GHC 8.0 is problematic
tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk
Sat Feb 6 12:31:57 UTC 2016
On Sat, Feb 06, 2016 at 01:27:00PM +0100, Ben Gamari wrote:
> Tom Ellis <tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk> writes:
> > On Fri, Feb 05, 2016 at 07:19:25PM +0000, Tom Ellis wrote:
> >> On Fri, Feb 05, 2016 at 01:13:23PM -0500, Richard Eisenberg wrote:
> >> > We're in a bit of a bind in all this. We really need the fancy type for
> >> > ($) 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'?
> 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)
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