В письме от 8 мая 2017 22:08:51 пользователь Fabien COELHO написал:
> > while (thread->throttle_trigger < now_us -
> > latency_limit &&
> >
> > /* with -t, do not overshoot */
> > (nxacts <= 0 || st->cnt < nxacts))
>
> ...
>
> > if (nxacts > 0 && st->cnt >= nxacts)
> > {
> >
> > st->state = CSTATE_FINISHED;
> > break;
> >
> > }
> >
st->cnt -- number of transactions finished successed or failed, right?
one iteration of for(;;) is for one transaction or really less. Right? We
can't process two tansactions in one iteration of this loop. So we can't
increase st->cnt more then once during one iteration?
So let's look at the while loop:
while (thread->throttle_trigger < now_us - latency_limit
&& /* with -t, do not overshoot */ (nxacts <= 0 || st->cnt < nxacts))
{ processXactStats(thread, st, &now, true, agg); /* next
rendez-vous*/ wait = getPoissonRand(thread, throttle_delay);
thread->throttle_trigger+= wait; st->txn_scheduled = thread->throttle_trigger;
}
Let's imagine that thread->throttle_trigger is now_us - 10 000,
latency_limit is 5 000 and throttle_delay is 100
How many times you would call processXactStats in this while loop?
And each time it would do st->cnt++
And this while loop is inside for(;;) in which as I said above we can do st-
>cnt++ not more than once. I see no logic here.
PS This is a fast reply. May be it will make things clear fast wither for me
or for you. I will carefully answer your full letter tomorrow (I hope nothing
will prevent me from doing it)
--
Nikolay Shaplov, independent Perl & C/C++ developer. Available for hire.
