User threads are interrupted uninterruptibly (see /proc/interrupts) and
pinning your thread/process to a particular CPU
does not guarantee that the OS won't run other threads on the same CPU,
just that yours won't be run on others.
More importantly, however, in this case is the fact that System.nanoTime()
is used to keep the CPU busy.
System.nanoTime() translates to clock_gettime(CLOCK_MONOTONIC) and that
thing is far more complex than
just 'rdtsc' and a couple of arithmetic operations. For one thing, it
occasionally makes system calls.
Not only those system calls take much longer to execute, you relinquish
control to the kernel at a
well defined point so it may decide to run stuff it wouldn't during
scheduling interrupts.
Anyway, all that to support a simple idea that after 5 seconds of
"spinning" in System.nanoTime()
you can be sure that cached 'calculation', 'rand' and whatever 'rand' uses
internally are long gone.
That is to say that when you measure your computation for the first time
after busyPause()
you measure, among other things, how long it takes to read all those values
from lower levels of
the memory hierarchy. My one-liner was intended to show that if we
'prefetch' the values the
measured time goes down significantly.
Whatever other theory you may have, it now has to provide an account for
that change as well.
This is not a full solution to the mystery you started with but an
important factor not to be
neglected.
-- Oleg
On Saturday, May 13, 2017 at 8:03:22 PM UTC-7, J Crawford wrote:
>
>
> Hi Oleg,
>
> I made sure to use thread pinning so the thread was pinned to an isolated
> CPU core. Furthermore, because the thread is spinning, I wouldn't think it
> is being interrupted or released from the CPU core.
>
> Ivan from SO came up with a theory that you must also warmup dynamically
> the pauses, check it out: http://stackoverflow.com/a/43699146/7833248
>
> Also not sure what you are trying to accomplish / prove with that extra
> line after busyPause. Can you elaborate?
>
> Thanks!
>
> -JC
>
> On Saturday, May 13, 2017 at 2:04:55 AM UTC-5, Oleg Mazurov wrote:
>>
>> Here is my theory. After warmup, every call to busyPause() takes 5
>> seconds.
>> During that time the thread will be interrupted many times. Even if it is
>> run
>> on the same CPU after every interrupt, the cache state will be gone.
>> Basically, what you measure after busyPause() is repopulating your caches.
>> To test this theory, add one line after busyPause() which does that
>> proactively
>> before measurement:
>>
>> double x = busyPause(interval);
>> + calculation += x / (rand.nextInt(5) + 1);
>>
>>
>> -- Oleg
>>
>
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