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

23 Aug 2016

      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
?