[Haskell-cafe] Proposal: (.:) operator in base.

[Matthew Pickering]

I think the point which no-one has articulated yet is that the
source-level arity of (.) affects whether GHC will decide to inline
it.
Only fully saturated applications are inlined, thus by having two
explicit arguments GHC will inline (.) in the common case where is it
used with two arguments which will improve optimisation opportunities
for code which composes many functions.

On the other hand, it is not very common to write

> (.) f g x

which is why (.) is defined in such a way.

Matt





On Tue, Aug 23, 2016 at 8:33 AM, Tom Ellis
<tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk> wrote:
> On the contrary, they're exactly the same (on GHC 7.6.3):
>
>     module Foo where
>
>     comp1 f g x = f (g x)
>
>     comp2 f g = \x -> f (g x)
>
>
>     % ghc -O0 -dsuppress-all -fforce-recomp -no-link -ddump-prep test.hs
>     [1 of 1] Compiling Foo              ( test.hs, test.o )
>
>     ==================== CorePrep ====================
>     Result size of CorePrep = {terms: 24, types: 36, coercions: 0}
>
>     comp2
>     comp2 =
>       \ @ t_aeU @ t1_aeV @ t2_aeW f_sfz g_sfy x_sfx ->
>         let {
>           sat_sfI
>           sat_sfI = g_sfy x_sfx } in
>         f_sfz sat_sfI
>
>     comp1
>     comp1 =
>       \ @ t_af5 @ t1_af6 @ t2_af7 f_sfG g_sfF x_sfE ->
>         let {
>           sat_sfJ
>           sat_sfJ = g_sfF x_sfE } in
>         f_sfG sat_sfJ
>
> (the same holds for -O2, if you compile them separately)
>
> On Tue, Aug 23, 2016 at 05:19:38PM +1000, Ben wrote:
>> At the semantic level of "does my program compute correct results" they're
>> identical.  At the operational level of "how fast does my program run"
>> they're different.
>>
>>
>> On August 23, 2016 5:09:19 PM GMT+10:00, Tom Ellis <tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk> wrote:
>> >On Mon, Aug 22, 2016 at 10:23:07PM -0700, wren romano wrote:
>> >>     (.) f g = \x -> f (g x)
>> >>
>> >> vs:
>> >>
>> >>     (.) f g x = f (g x)
>> >>
>> >> has ramifications, though it's fairly easy to guess which one of
>> >those
>> >> two will be most performant.
>> >
>> >Are these not synonyms?  What is the meaning of
>> >
>> >    fargs var = expr
>> >
>> >if not
>> >
>> >    fargs = \var -> expr
>> >
>> >?
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