Re: Cleaning up the KVM clock

[Marcelo Tosatti <mtosatti@redhat.com>]

On Sat, Dec 20, 2014 at 07:31:19PM -0800, Andy Lutomirski wrote:
> I'm looking at the vdso timing code, and I'm puzzled by the pvclock
> code.  My motivation is comprehensibility, performance, and
> correctness.
> 
> # for i in `seq 10`; do ./timing_test_64 10 vclock_gettime 0; done
> 10000000 loops in 0.69138s = 69.14 nsec / loop
> 10000000 loops in 0.63614s = 63.61 nsec / loop
> 10000000 loops in 0.63213s = 63.21 nsec / loop
> 10000000 loops in 0.63087s = 63.09 nsec / loop
> 10000000 loops in 0.63079s = 63.08 nsec / loop
> 10000000 loops in 0.63096s = 63.10 nsec / loop
> 10000000 loops in 0.63096s = 63.10 nsec / loop
> 10000000 loops in 0.63062s = 63.06 nsec / loop
> 10000000 loops in 0.63100s = 63.10 nsec / loop
> 10000000 loops in 0.63112s = 63.11 nsec / loop
> bash-4.3# echo tsc
> >/sys/devices/system/clocksource/clocksource0/current_clocksource
> [   45.957524] Switched to clocksource tsc
> bash-4.3# for i in `seq 10`; do ./timing_test_64 10 vclock_gettime 0;
> done10000000 loops in 0.33583s = 33.58 nsec / loop
> 10000000 loops in 0.28530s = 28.53 nsec / loop
> 10000000 loops in 0.28904s = 28.90 nsec / loop
> 10000000 loops in 0.29001s = 29.00 nsec / loop
> 10000000 loops in 0.28775s = 28.78 nsec / loop
> 10000000 loops in 0.30102s = 30.10 nsec / loop
> 10000000 loops in 0.28006s = 28.01 nsec / loop
> 10000000 loops in 0.28584s = 28.58 nsec / loop
> 10000000 loops in 0.28175s = 28.17 nsec / loop
> 10000000 loops in 0.28724s = 28.72 nsec / loop
> 
> The current code is rather slow, especially compared to the tsc variant.
> 
> The algorithm used by the pvclock vgetsns implementation is, approximately:
> 
> cpu = getcpu;
> pvti = pointer to the relevant paravirt data
> version = pvti->version;
> rdtsc_barrier();
> tsc = rdtsc()
> delta = (tsc - x) * y >> z;
> cycles = delta + w;
> flags = pvti->flags;
> rdtsc_barrier();  <-- totally unnecessary

Why?

"The RDTSC instruction is not a serializing instruction. It does not
necessarily wait until all previous instructions have been executed
before reading the counter. Similarly, subsequent instructions may begin
execution before the read operation is performed."