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.
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