Re: [PATCH] Guest system time jumps when new vCPUs is hot-added

[Zelin Deng]

Got it. Many thanks, Thomas.

On 2021/4/30 上午12:02, Thomas Gleixner wrote:

> On Thu, Apr 29 2021 at 17:38, Zelin Deng wrote:
> > On 2021/4/29 下午4:46, Thomas Gleixner wrote:
> > > And that validation expects that the CPUs involved run in a tight loop
> > > concurrently so the TSC readouts which happen on both can be reliably
> > > compared.
> > >
> > > But this cannot be guaranteed on vCPUs at all, because the host can
> > > schedule out one or both at any point during that synchronization
> > > check.
> > Is there any plan to fix this?
> The above cannot be fixed.
>
> As I said before the solution is:
>
> > > A two socket guest setup needs to have information from the host that
> > > TSC is usable and that the socket sync check can be skipped. Anything
> > > else is just doomed to fail in hard to diagnose ways.
> > Yes, I had tried to add "tsc=unstable" to skip tsc sync.  However if a
> tsc=unstable? Oh well.
>
> > user process which is not pined to vCPU is using rdtsc, it can get tsc
> > warp, because it can be scheduled among vCPUs.  Does it mean user
> Only if the hypervisor is not doing the right thing and makes sure that
> all vCPUs have the same tsc offset vs. the host TSC.
>
> > applications have to guarantee itself to use rdtsc only when TSC is
> > reliable?
> If the TSCs of CPUs are not in sync then the kernel does the right thing
> and uses some other clocksource for the various time interfaces, e.g.
> the kernel provides clock_getttime() which guarantees to be correct
> whether TSC is usable or not.
>
> Any application using RDTSC directly is own their own and it's not a
> kernel problem.
>
> The host kernel cannot make guarantees that the hardware is sane neither
> can a guest kernel make guarantees that the hypervisor is sane.
>
> Thanks,
>
>          tglx