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Date:   Thu, 5 Jul 2018 15:42:14 +0200
From:   Peter Zijlstra <peterz@infradead.org>
To:     Xunlei Pang <xlpang@linux.alibaba.com>
Cc:     Ingo Molnar <mingo@redhat.com>,
        Frederic Weisbecker <frederic@kernel.org>,
        Tejun Heo <tj@kernel.org>, linux-kernel@vger.kernel.org
Subject: Re: [PATCH] sched/cputime: Ensure correct utime and stime proportion
Message-ID: <20180705134214.GF2494@hirez.programming.kicks-ass.net>
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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
!VIRT_CPU_ACCOUNTING.


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