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

[Tom Ellis]

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
> >
> >?