linux-kernel - Re: [RFC] change non-atomic bitops method

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Message-Id: <20150202223851.f30768d0.akpm@linux-foundation.org>
Date:	Mon, 2 Feb 2015 22:38:51 -0800
From:	Andrew Morton <akpm@...ux-foundation.org>
To:	"Wang, Yalin" <Yalin.Wang@...ymobile.com>
Cc:	"'Kirill A. Shutemov'" <kirill@...temov.name>,
	"'arnd@...db.de'" <arnd@...db.de>,
	"'linux-arch@...r.kernel.org'" <linux-arch@...r.kernel.org>,
	"'linux-kernel@...r.kernel.org'" <linux-kernel@...r.kernel.org>,
	"'linux@....linux.org.uk'" <linux@....linux.org.uk>,
	"'linux-arm-kernel@...ts.infradead.org'" 
	<linux-arm-kernel@...ts.infradead.org>
Subject: Re: [RFC] change non-atomic bitops method

On Tue, 3 Feb 2015 13:42:45 +0800 "Wang, Yalin" <Yalin.Wang@...ymobile.com> wrote:
>
> ...
>
> #ifdef CHECK_BEFORE_SET
> 			if (p[i] != times)
> #endif
>
> ...
>
> ----
> One run on CPU0, reader thread run on CPU1,
> Test result:
> sudo ./cache_test
> reader:8.426228173
> 8.672198335
> 
> With -DCHECK_BEFORE_SET
> sudo ./cache_test_check
> reader:7.537036819
> 10.799746531
> 

You aren't measuring the right thing.  You should compare

	if (p[i] != x)
		p[i] = x;

versus

	p[i] = x;

and you should do this for two cases:

a) p[i] == x

b) p[i] != x


The first code sequence will be slower when (p[i] != x) and faster when
(p[i] == x).


Next, we should instrument the kernel to work out the frequency of
set_bit on an already-set bit.

It is only with both these ratios that we can work out whether the
patch is a net gain.  My suspicion is that set_bit on an already-set
bit is so rare that the patch will be a loss.
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