Re: [RFC 2/2] x86, vdso, pvclock: Simplify and speed up the vdso pvclock reader

Andy Lutomirski
2015-01-05T18:56:29Z

On Mon, Jan 5, 2015 at 7:25 AM, Marcelo Tosatti <mtosatti@redhat.com> wrote:
> On Mon, Dec 22, 2014 at 04:39:57PM -0800, Andy Lutomirski wrote:
>> The pvclock vdso code was too abstracted to understand easily and
>> excessively paranoid. Simplify it for a huge speedup.
>>
>> This opens the door for additional simplifications, as the vdso no
>> longer accesses the pvti for any vcpu other than vcpu 0.
>>
>> Before, vclock_gettime using kvm-clock took about 64ns on my machine.
>> With this change, it takes 19ns, which is almost as fast as the pure TSC
>> implementation.
>>
>> Signed-off-by: Andy Lutomirski <luto@amacapital.net>
>> ---
>> arch/x86/vdso/vclock_gettime.c | 82 ++++++++++++++++++++++++------------------
>> 1 file changed, 47 insertions(+), 35 deletions(-)
>>
>> diff --git a/arch/x86/vdso/vclock_gettime.c b/arch/x86/vdso/vclock_gettime.c
>> index 9793322751e0..f2e0396d5629 100644
>> --- a/arch/x86/vdso/vclock_gettime.c
>> +++ b/arch/x86/vdso/vclock_gettime.c
>> @@ -78,47 +78,59 @@ static notrace const struct pvclock_vsyscall_time_info *get_pvti(int cpu)
>>
>> static notrace cycle_t vread_pvclock(int *mode)
>> {
>> - const struct pvclock_vsyscall_time_info *pvti;
>> + const struct pvclock_vcpu_time_info *pvti = &get_pvti(0)->pvti;
>> cycle_t ret;
>> - u64 last;
>> - u32 version;
>> - u8 flags;
>> - unsigned cpu, cpu1;
>> -
>> + u64 tsc, pvti_tsc;
>> + u64 last, delta, pvti_system_time;
>> + u32 version, pvti_tsc_to_system_mul, pvti_tsc_shift;
>>
>> /*
>> - * Note: hypervisor must guarantee that:
>> - * 1. cpu ID number maps 1:1 to per-CPU pvclock time info.
>> - * 2. that per-CPU pvclock time info is updated if the
>> - * underlying CPU changes.
>> - * 3. that version is increased whenever underlying CPU
>> - * changes.
>> + * Note: The kernel and hypervisor must guarantee that cpu ID
>> + * number maps 1:1 to per-CPU pvclock time info.
>> + *
>> + * Because the hypervisor is entirely unaware of guest userspace
>> + * preemption, it cannot guarantee that per-CPU pvclock time
>> + * info is updated if the underlying CPU changes or that that
>> + * version is increased whenever underlying CPU changes.
>> + *
>> + * On KVM, we are guaranteed that pvti updates for any vCPU are
>> + * atomic as seen by *all* vCPUs. This is an even stronger
>> + * guarantee than we get with a normal seqlock.
>> *
>> + * On Xen, we don't appear to have that guarantee, but Xen still
>> + * supplies a valid seqlock using the version field.
>> +
>> + * We only do pvclock vdso timing at all if
>> + * PVCLOCK_TSC_STABLE_BIT is set, and we interpret that bit to
>> + * mean that all vCPUs have matching pvti and that the TSC is
>> + * synced, so we can just look at vCPU 0's pvti.
>> */
>
> Can Xen guarantee that ?

I think so, vacuously. Xen doesn't seem to set PVCLOCK_TSC_STABLE_BIT
at all. I have no idea going forward, though.

Xen people?

>
>> - do {
>> - cpu = __getcpu() & VGETCPU_CPU_MASK;
>> - /* TODO: We can put vcpu id into higher bits of pvti.version.
>> - * This will save a couple of cycles by getting rid of
>> - * __getcpu() calls (Gleb).
>> - */
>> -
>> - pvti = get_pvti(cpu);
>> -
>> - version = __pvclock_read_cycles(&pvti->pvti, &ret, &flags);
>> -
>> - /*
>> - * Test we're still on the cpu as well as the version.
>> - * We could have been migrated just after the first
>> - * vgetcpu but before fetching the version, so we
>> - * wouldn't notice a version change.
>> - */
>> - cpu1 = __getcpu() & VGETCPU_CPU_MASK;
>> - } while (unlikely(cpu != cpu1 ||
>> - (pvti->pvti.version & 1) ||
>> - pvti->pvti.version != version));
>> -
>> - if (unlikely(!(flags & PVCLOCK_TSC_STABLE_BIT)))
>> +
>> + if (unlikely(!(pvti->flags & PVCLOCK_TSC_STABLE_BIT))) {
>> *mode = VCLOCK_NONE;
>> + return 0;
>> + }
>
> This check must be performed after reading a stable pvti.
>

We can even read it in the middle, guarded by the version checks.
I'll do that for v2.

>> +
>> + do {
>> + version = pvti->version;
>> +
>> + /* This is also a read barrier, so we'll read version first. */
>> + rdtsc_barrier();
>> + tsc = __native_read_tsc();
>> +
>> + pvti_tsc_to_system_mul = pvti->tsc_to_system_mul;
>> + pvti_tsc_shift = pvti->tsc_shift;
>> + pvti_system_time = pvti->system_time;
>> + pvti_tsc = pvti->tsc_timestamp;
>> +
>> + /* Make sure that the version double-check is last. */
>> + smp_rmb();
>> + } while (unlikely((version & 1) || version != pvti->version));
>> +
>> + delta = tsc - pvti_tsc;
>> + ret = pvti_system_time +
>> + pvclock_scale_delta(delta, pvti_tsc_to_system_mul,
>> + pvti_tsc_shift);
>
> The following is possible:
>
> 1) State: all pvtis marked as PVCLOCK_TSC_STABLE_BIT.
> 1) Update request for all vcpus, for a TSC_STABLE_BIT -> ~TSC_STABLE_BIT
> transition.
> 2) vCPU-1 updates its pvti with new values.
> 3) vCPU-0 still has not updated its pvti with new values.
> 4) vCPU-1 VM-enters, uses vCPU-0 values, even though it has been
> notified of a TSC_STABLE_BIT -> ~TSC_STABLE_BIT transition.
>
> The update is not actually atomic across all vCPUs, its atomic in
> the sense of not allowing visibility of distinct
> system_timestamp/tsc_timestamp values.
>

Hmm. In step 4, is there a guarantee that vCPU-0 won't VM-enter until
it gets marked unstable? Otherwise the vdso could could just as
easily be called from vCPU-1, migrated to vCPU-0, read the data
complete with stale stable bit, and get migrated back to vCPU-1.

But I thought that KVM currently froze all vCPUs when updating pvti
for any of them. How can this happen? I admit I don't really
understand the update request code.

--Andy

--
Andy Lutomirski
AMA Capital Management, LLC