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