Re: [RFC 2/2] x86, vdso, pvclock: Simplify and speed up the vdso pvclock reader
From: Andy Lutomirski
Date: Wed Jan 07 2015 - 02:18:48 EST
On Tue, Jan 6, 2015 at 9:38 PM, Paolo Bonzini <pbonzini@xxxxxxxxxx> wrote:
>
>
> On 06/01/2015 17:56, Andy Lutomirski wrote:
>> Still no good. We can migrate a bunch of times so we see the same CPU
>> all three times
>
> There are no three times. The CPU you see here:
>
>>>
>>>
>>> // ... compute nanoseconds from pvti and tsc ...
>>> rmb();
>>> } while(v != pvti->version);
>
> is the same you read here:
>
>>> cpu = get_cpu();
>
> The algorithm is:
I still don't see why this is safe, and I think that the issue is that
you left out part of the loop.
>
> 1) get a consistent (cpu, version, tsc)
>
> 1.a) get cpu
Suppose we observe cpu 0.
> 1.b) get pvti[cpu]->version, ignoring low bit
Missing step, presumably here: read pvti[cpu]->tsc_timestamp, scale,
etc. This could all execute on vCPU 1. We could read values that are
inconsistent with each other.
> 1.c) get (tsc, cpu)
Now we could end up back on vCPU 0.
> 1.d) if cpu from 1.a and 1.c do not match, loop
> 1.e) if pvti[cpu] was being updated, we'll loop later
>
> 2) compute nanoseconds from pvti[cpu] and tsc
>
> 3) if pvti[cpu] changed under our feet during (2), i.e. version doesn't
> match, retry.
>
> As long as the CPU is consistent between get_cpu() and rdtscp(), there
> is no problem with migration, because pvti is always accessed for that
> one CPU. If there were any problem, it would be caught by the version
> check. Writing it down with two nested do...whiles makes it clearer IMHO.
Why exactly would it be caught by the version check?
My ugly patch tries to make the argument that, at any point at which
we observe ourselves to be on a given vCPU, that vCPU won't be
updating pvti. That means that, if version doesn't change between two
consecutive observations that we're on that vCPU, then we're okay.
This IMO sucks. It's fragile, it's hard to make a coherent argument
about correctness, and it requires at least two getcpu-like operations
to read the time. Those operations are *slow*. One is much better
than two, and zero is much better than one.
>
>> and *still* don't get a consistent read, unless we
>> play nasty games with lots of version checks (I have a patch for that,
>> but I don't like it very much). The patch is here:
>>
>> https://git.kernel.org/cgit/linux/kernel/git/luto/linux.git/commit/?h=x86/vdso_paranoia&id=a69754dc5ff33f5187162b5338854ad23dd7be8d
>>
>> but I don't like it.
>>
>> Thus far, I've been told unambiguously that a guest can't observe pvti
>> while it's being written, and I think you're now telling me that this
>> isn't true and that a guest *can* observe pvti while it's being
>> written while the low bit of the version field is not set. If so,
>> this is rather strongly incompatible with the spec in the KVM docs.
>
> Where am I saying that?
I thought the conclusion from what you and Marcelo pointed out about
the code was that, once the first vCPU updated its pvti, it could
start running guest code while the other vCPUs are still updating
pvti, so its guest code can observe the other vCPUs mid-update.
>> Also, if you do this, can you also make setting and clearing
>> STABLE_BIT properly atomic across all vCPUs? Or at least do something
>> like setting it last and clearing it first on vPCU 0?
>
> That would be nice if you want to make the pvclock area fit in a single
> page. However, it would have to be a separate flag bit, or a separate
> CPUID feature.
It would be nice to have. Although I think that fixing the host to
increment version pre-update and post-update may actually be good
enough. Is there any case in which it would fail in practice if we
made that fix and always looked at pvti 0?
--Andy
>
> Paolo
--
Andy Lutomirski
AMA Capital Management, LLC
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