Re: [PATCH 1/1] cputime: Make the reported utime+stime correspond to the actual runtime.

From: Fredrik MarkstrÃm
Date: Tue Jul 07 2015 - 08:11:38 EST


Just to let you know, I've tested your last patch and it solves all my
original problems (which is should since the code is functionally
equivalent).

/Fredrik

On Tue, Jul 7, 2015 at 10:09 AM, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
> On Tue, Jul 07, 2015 at 09:59:54AM +0200, Peter Zijlstra wrote:
>> > > + /*
>> > > + * Make sure stime doesn't go backwards; this preserves monotonicity
>> > > + * for utime because rtime is monotonic.
>> > > + *
>> > > + * utime_i+1 = rtime_i+1 - stime_i
>> > > + * = rtime_i+1 - (rtime_i - stime_i)
>> > > + * = (rtime_i+1 - rtime_i) + stime_i
>> > > + * >= stime_i
>> > > + */
>
> Argh, just noticed I messed that up, it should read:
>
> + /*
> + * Make sure stime doesn't go backwards; this preserves monotonicity
> + * for utime because rtime is monotonic.
> + *
> + * utime_i+1 = rtime_i+1 - stime_i
> + * = rtime_i+1 - (rtime_i - utime_i)
> + * = (rtime_i+1 - rtime_i) + utime_i
> + * >= utime_i
> + */
>
> I got some [us] confusion. Typing is hard.
>
> So we compute: utime = rtime - stime, which we'll denote as:
>
> utime_i+1 = rtime_i+1 - stime_i
>
> since: stime_i + utime_i = rtime_i, we can write: stime_i = rtime_i -
> utime_i and substitute in the above:
>
> = rtime_i+1 - (rtime_i - utime_i)
>
> Rearrange terms:
>
> = (rtime_i+1 - rtime_i) + utime_i
>
> And since we have: rtime_i+1 >= rtime_i, which we can write like:
> rtime_i+1 - rtime_i >= 0, substitution gives:
>
> >= utime_i
>
> And we've proven that the new utime must be equal or greater than the
> previous utime, because rtime is monotonic.



--
/Fredrik
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