On 2/29/08, Peter Zijlstra <[EMAIL PROTECTED]> 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? ------------------------------------------------------------------------- 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/ _______________________________________________ kvm-devel mailing list kvm-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/kvm-devel
