And also sometimes wrong. I don't really understand why, but sometimes eta *expansion* is important when working around RULES. On Aug 23, 2016 5:42 AM, "Alan & Kim Zimmerman" wrote: Suddenly the eta reduce hlint warning becomes much more important. Alan On 23 Aug 2016 11:22 a.m., "Tom Ellis" wrote:

*Very* interesting. Here is the key section of the GHC users guide:

GHC will only inline the function if it is fully applied, where “fully applied” means applied to as many arguments as appear (syntactically) on the LHS of the function definition

That this is a good heuristic is *extremely* counterintuitive to me. I would have supposed that being more explicit, for example

{-# INLINE (.) f g #-}

to inline on all static applications of at least two arguments would have been a much clearer way to communicate this message.

What's the rationale behind the current behaviour?

Tom

...

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 [...]

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"

On Tue, Aug 23, 2016 at 09:56:25AM +0100, Matthew Pickering wrote: 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@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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