> -----Original Message-----
> From: Andrew Morton [mailto:a...@linux-foundation.org]
> Sent: Tuesday, February 03, 2015 2:39 PM
> To: Wang, Yalin
> Cc: 'Kirill A. Shutemov'; 'a...@arndb.de'; 'linux-a...@vger.kernel.org';
> 'linux-kernel@vger.kernel.org'; 'li...@arm.linux.org.uk'; 'linux-arm-
> ker...@lists.infradead.org'
> Subject: Re: [RFC] change non-atomic bitops method
>
> On Tue, 3 Feb 2015 13:42:45 +0800 "Wang, Yalin" <yalin.w...@sonymobile.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.
I see, let's change the test a little:
1)
memset(p, 0, SIZE);
if (p[i] != 0)
p[i] = 0; // never called
#sudo ./cache_test_check
6.698153838
reader:7.529402625
2)
memset(p, 0, SIZE);
if (p[i] == 0)
p[i] = 0; // always called
#sudo ./cache_test_check
reader:7.895421311
9.000889973
Thanks
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