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Re: block-sha1: improve code on large-register-set machines

From
Linus Torvalds <torvalds@linux-foundation.org>
Date
Aug 11, 2009, 15:43 UTC
Message-ID
<alpine.LFD.2.01.0908110810410.3417@localhost.localdomain>
In-Reply-To
<alpine.LFD.2.00.0908102246210.10633@xanadu.home>
On Tue, 11 Aug 2009, Nicolas Pitre wrote:
Show 8 quoted lines
> 
> BLK_SHA1:	 5.280s		[original]
> BLK_SHA1:	 7.410s		[with SMALL_REGISTER_SET defined]
> BLK_SHA1:	 7.480s		[with 'W(x)=(val);asm("":"+m" (W(x)))']
> BLK_SHA1:	 4.980s		[with 'W(x)=(val);asm("":::"memory")']
> 
> At this point the generated assembly is pretty slick.  I bet the full 
> memory barrier might help on x86 as well.

No, I had tested that earlier - single-word memory barrier for some reason gets _much_ better numbers at least on x86-64. We're talking

	linus            1.46       418.2
vs
	linus           2.004       304.6
kind of differences. With the "+m" it outperforms openssl (375-380MB/s).

The "volatile unsigned int *" cast looks pretty much like the "+m" version to me, but Arthur got a speedup from whatever gcc code generation differences on his P4.

The really fundamental and basic problem with gcc on this code is that gcc does not see _any_ difference what-so-ever between the five variables declared with

	unsigned int A, B, C, D, E;
and the sixteen variables declared with
	unsigned int array[16];

and considers those all to be 21 local variables. It really seems to think that they are all 100% equivalent, and gcc totally ignores me doing things like adding "register" to the A-E ones etc.

And if you are a compiler, and think that the routine has 21 equal register variables, you're going to do crazy reload sh*t when you have only 7 (or 15) GP registers. So doing that full memory barrier seems to just take that random situation, and force some random variable to be spilled (this is all from looking at the generated code, not from looking at gcc).

In contrast, with the _targeted_ thing ("you'd better write back into array[]") we force gcc to spill the array[16] values, and not the A-E ones, and that's why it seems to make such a big difference.

And no, I'm not sure why ARM apparently doesn't show the same behavior. Or maybe it does, but with an in-order core it doesn't matter as much which registers you keep reloading - you'll be serialized all the time _anyway_.

			Linus
Previous: Brandon CaseyNext: Nicolas Pitre
Message 11 of 16 in “block-sha1: improve code on large-register-set machines”
  1. Linus TorvaldsAug 10, 2009
  2. Nicolas PitreAug 11, 2009
  3. Linus TorvaldsAug 11, 2009
  4. Nicolas PitreAug 11, 2009
  5. Nicolas PitreAug 11, 2009
  6. Brandon CaseyAug 11, 2009
  7. Nicolas PitreAug 11, 2009
  8. Brandon CaseyAug 11, 2009
  9. Linus TorvaldsAug 11, 2009
  10. Brandon CaseyAug 11, 2009
  11. Linus TorvaldsAug 11, 2009
  12. Nicolas PitreAug 11, 2009
  13. Linus TorvaldsAug 11, 2009
  14. Linus TorvaldsAug 11, 2009
  15. Nicolas PitreAug 12, 2009
  16. Artur SkawinaAug 11, 2009

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