On Mon, 2016-06-06 at 15:40 +0200, Paolo Bonzini wrote: > > On 02/06/2016 15:59, Rik van Riel wrote: > > > > If a guest is saved to disk and later restored (eg. after > > a host reboot), or live migrated to another host, I would > > expect to get totally disjoint steal time statistics from > > the "new run" of the guest (which is the same run of the > > guest OS). > Why? The preexisting guest steal time is always added to by > KVM, so the time won't restart from zero. > > Continuing the previous count on CPU hot-unplug followed by hot-plug > is less obvious, but I think it's overall the right thing to do. > > In fact, I was going to test a patch this week as simple as this: > > diff --git a/arch/x86/kernel/kvm.c b/arch/x86/kernel/kvm.c > index eea2a6f72b31..1ef5e48b3a36 100644 > --- a/arch/x86/kernel/kvm.c > +++ b/arch/x86/kernel/kvm.c > @@ -301,8 +301,6 @@ static void kvm_register_steal_time(void) > if (!has_steal_clock) > return; > > - memset(st, 0, sizeof(*st)); > - > wrmsrl(MSR_KVM_STEAL_TIME, (slow_virt_to_phys(st) | > KVM_MSR_ENABLED)); By removing the memset from initial bootup allocation, can't that cause the steal time to "increase by a ludicrous amount" the very first time it is compared with the arch independent value in the scheduler code? In other words, would removal of the memset result in still requiring Wanpeng's patch? What am I overlooking? Is there something preventing a non-zero value right at the beginning? Also, is there a chance of ending up with a non-zero bit in the seqcount if the memset is removed? -- All Rights Reversed.
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