* Can Linux kernel handle unsynced TSC?
@ 2008-02-29 8:55 ` Zhao Forrest
0 siblings, 0 replies; 9+ messages in thread
From: Zhao Forrest @ 2008-02-29 8:55 UTC (permalink / raw)
To: linux-kernel; +Cc: kvm-devel
Sorry for reposting it.
For example,
1 rdtsc() is invoked on CPU0
2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
that the second rdtsc() doesn't return a value smaller than the one
returned by the first rdtsc()?
Thanks,
Forrest
^ permalink raw reply [flat|nested] 9+ messages in thread
* Can Linux kernel handle unsynced TSC?
@ 2008-02-29 8:55 ` Zhao Forrest
0 siblings, 0 replies; 9+ messages in thread
From: Zhao Forrest @ 2008-02-29 8:55 UTC (permalink / raw)
To: linux-kernel; +Cc: kvm-devel
Sorry for reposting it.
For example,
1 rdtsc() is invoked on CPU0
2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
that the second rdtsc() doesn't return a value smaller than the one
returned by the first rdtsc()?
Thanks,
Forrest
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^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Can Linux kernel handle unsynced TSC?
2008-02-29 8:55 ` Zhao Forrest
@ 2008-02-29 10:43 ` Peter Zijlstra
-1 siblings, 0 replies; 9+ messages in thread
From: Peter Zijlstra @ 2008-02-29 10:43 UTC (permalink / raw)
To: Zhao Forrest; +Cc: linux-kernel, kvm-devel
On Fri, 2008-02-29 at 16:55 +0800, Zhao Forrest wrote:
> Sorry for reposting it.
>
> For example,
> 1 rdtsc() is invoked on CPU0
> 2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
> 3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
> that the second rdtsc() doesn't return a value smaller than the one
> returned by the first rdtsc()?
No, rdtsc() goes directly to the hardware. You need a (preferably cheap)
clock abstraction layer on top if you need this.
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Can Linux kernel handle unsynced TSC?
@ 2008-02-29 10:43 ` Peter Zijlstra
0 siblings, 0 replies; 9+ messages in thread
From: Peter Zijlstra @ 2008-02-29 10:43 UTC (permalink / raw)
To: Zhao Forrest; +Cc: kvm-devel, linux-kernel
On Fri, 2008-02-29 at 16:55 +0800, Zhao Forrest wrote:
> Sorry for reposting it.
>
> For example,
> 1 rdtsc() is invoked on CPU0
> 2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
> 3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
> that the second rdtsc() doesn't return a value smaller than the one
> returned by the first rdtsc()?
No, rdtsc() goes directly to the hardware. You need a (preferably cheap)
clock abstraction layer on top if you need this.
-------------------------------------------------------------------------
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^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Can Linux kernel handle unsynced TSC?
2008-02-29 10:43 ` Peter Zijlstra
@ 2008-02-29 14:20 ` Zhao Forrest
-1 siblings, 0 replies; 9+ messages in thread
From: Zhao Forrest @ 2008-02-29 14:20 UTC (permalink / raw)
To: Peter Zijlstra; +Cc: linux-kernel, kvm-devel
On 2/29/08, Peter Zijlstra <peterz@infradead.org> wrote:
>
> On Fri, 2008-02-29 at 16:55 +0800, Zhao Forrest wrote:
> > Sorry for reposting it.
> >
> > For example,
> > 1 rdtsc() is invoked on CPU0
> > 2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
> > 3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
> > that the second rdtsc() doesn't return a value smaller than the one
> > returned by the first rdtsc()?
>
> No, rdtsc() goes directly to the hardware. You need a (preferably cheap)
> clock abstraction layer on top if you need this.
Thank you for the clarification. I think gettimeofday() is such kind
of clock abstraction layer, am I right?
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Can Linux kernel handle unsynced TSC?
@ 2008-02-29 14:20 ` Zhao Forrest
0 siblings, 0 replies; 9+ messages in thread
From: Zhao Forrest @ 2008-02-29 14:20 UTC (permalink / raw)
To: Peter Zijlstra; +Cc: kvm-devel, linux-kernel
On 2/29/08, Peter Zijlstra <peterz@infradead.org> wrote:
>
> On Fri, 2008-02-29 at 16:55 +0800, Zhao Forrest wrote:
> > Sorry for reposting it.
> >
> > For example,
> > 1 rdtsc() is invoked on CPU0
> > 2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
> > 3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
> > that the second rdtsc() doesn't return a value smaller than the one
> > returned by the first rdtsc()?
>
> No, rdtsc() goes directly to the hardware. You need a (preferably cheap)
> clock abstraction layer on top if you need this.
Thank you for the clarification. I think gettimeofday() is such kind
of clock abstraction layer, am I right?
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^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Can Linux kernel handle unsynced TSC?
2008-02-29 14:20 ` Zhao Forrest
@ 2008-02-29 14:27 ` Peter Zijlstra
-1 siblings, 0 replies; 9+ messages in thread
From: Peter Zijlstra @ 2008-02-29 14:27 UTC (permalink / raw)
To: Zhao Forrest; +Cc: linux-kernel, kvm-devel
On Fri, 2008-02-29 at 22:20 +0800, Zhao Forrest wrote:
> On 2/29/08, Peter Zijlstra <peterz@infradead.org> wrote:
> >
> > On Fri, 2008-02-29 at 16:55 +0800, Zhao Forrest wrote:
> > > Sorry for reposting it.
> > >
> > > For example,
> > > 1 rdtsc() is invoked on CPU0
> > > 2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
> > > 3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
> > > that the second rdtsc() doesn't return a value smaller than the one
> > > returned by the first rdtsc()?
> >
> > No, rdtsc() goes directly to the hardware. You need a (preferably cheap)
> > clock abstraction layer on top if you need this.
>
> Thank you for the clarification. I think gettimeofday() is such kind
> of clock abstraction layer, am I right?
Yes, gtod is one such a layer, however it fails the 'cheap' test for
many definitions of cheap.
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Can Linux kernel handle unsynced TSC?
@ 2008-02-29 14:27 ` Peter Zijlstra
0 siblings, 0 replies; 9+ messages in thread
From: Peter Zijlstra @ 2008-02-29 14:27 UTC (permalink / raw)
To: Zhao Forrest; +Cc: kvm-devel, linux-kernel
On Fri, 2008-02-29 at 22:20 +0800, Zhao Forrest wrote:
> On 2/29/08, Peter Zijlstra <peterz@infradead.org> wrote:
> >
> > On Fri, 2008-02-29 at 16:55 +0800, Zhao Forrest wrote:
> > > Sorry for reposting it.
> > >
> > > For example,
> > > 1 rdtsc() is invoked on CPU0
> > > 2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
> > > 3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
> > > that the second rdtsc() doesn't return a value smaller than the one
> > > returned by the first rdtsc()?
> >
> > No, rdtsc() goes directly to the hardware. You need a (preferably cheap)
> > clock abstraction layer on top if you need this.
>
> Thank you for the clarification. I think gettimeofday() is such kind
> of clock abstraction layer, am I right?
Yes, gtod is one such a layer, however it fails the 'cheap' test for
many definitions of cheap.
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^ permalink raw reply [flat|nested] 9+ messages in thread
* Can Linux kernel handle unsynced TSC?
@ 2008-02-29 8:43 Zhao Forrest
0 siblings, 0 replies; 9+ messages in thread
From: Zhao Forrest @ 2008-02-29 8:43 UTC (permalink / raw)
To: lkml; +Cc: kvm-devel
For example,
1 rdtsc() is invoked on CPU0
2 process is migrated to CPU1, and rdtsc() is invoked on CPU1
3 if TSC on CPU1 is slower than TSC on CPU0, can kernel guarantee
that the second rdtsc() doesn't return a value smaller than the one
returned by the first rdtsc()?
Thanks,
Forrest
-------------------------------------------------------------------------
This SF.net email is sponsored by: Microsoft
Defy all challenges. Microsoft(R) Visual Studio 2008.
http://clk.atdmt.com/MRT/go/vse0120000070mrt/direct/01/
^ permalink raw reply [flat|nested] 9+ messages in thread
end of thread, other threads:[~2008-02-29 14:27 UTC | newest]
Thread overview: 9+ messages (download: mbox.gz follow: Atom feed
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2008-02-29 8:55 Can Linux kernel handle unsynced TSC? Zhao Forrest
2008-02-29 8:55 ` Zhao Forrest
2008-02-29 10:43 ` Peter Zijlstra
2008-02-29 10:43 ` Peter Zijlstra
2008-02-29 14:20 ` Zhao Forrest
2008-02-29 14:20 ` Zhao Forrest
2008-02-29 14:27 ` Peter Zijlstra
2008-02-29 14:27 ` Peter Zijlstra
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2008-02-29 8:43 Zhao Forrest
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