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. ------------------------------------------------------------------------- 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
