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

23 Aug 2016

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

Matthew Pickering