Re: [PATCH] sched/cputime: Ensure correct utime and stime proportion

From: Peter Zijlstra
Date: Thu Jul 05 2018 - 09:42:29 EST

On Thu, Jul 05, 2018 at 09:21:15PM +0800, Xunlei Pang wrote:
> On 7/5/18 6:46 PM, Peter Zijlstra wrote:
> > On Wed, Jun 27, 2018 at 08:22:42PM +0800, Xunlei Pang wrote:
> >> tick-based whole utime is utime_0, tick-based whole stime
> >> is stime_0, scheduler time is rtime_0.
> >
> >> For a long time, the process runs mainly in userspace with
> >> run-sleep patterns, and because two different clocks, it
> >> is possible to have the following condition:
> >> rtime_0 < utime_0 (as with little stime_0)
> >
> > I don't follow... what?
> >
> > Why are you, and why do you think it makes sense to, compare rtime_0
> > against utime_0 ?
> >
> > The [us]time_0 are, per your earlier definition, ticks. They're not an
> > actual measure of time. Do not compare the two, that makes no bloody
> > sense.
> >
> [us]time_0 is task_struct:utime{stime}, I cited directly from
> cputime_adjust(), both in nanoseconds. I assumed "rtime_0 < utime_0"
> here to simple the following proof to help explain the problem we met.

In the !VIRT_CPU_ACCOUNTING case they (task_struct::[us]time) are not
actual durations. Yes, the happen to be accounted in multiples of
TICK_NSEC and thereby happen to carry a [ns] unit, but they are not
durations, they are samples.

(we just happen to store them in a [ns] unit because for
VIRT_CPU_ACCOUNTING they are in fact durations)

If 'rtime < utime' is not a valid assumption to build a problem on for

So please try again, so far you're not making any sense.