Messages in this thread | ![/](/images/icornerl.gif) | | Subject | Re: [RFC][PATCH 02/11] sched: SCHED_DEADLINE policy implementation. | From | Peter Zijlstra <> | Date | Thu, 15 Apr 2010 09:34:06 +0200 |
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On Tue, 2010-04-13 at 14:55 -0400, Steven Rostedt wrote: > On Tue, 2010-04-13 at 20:22 +0200, Peter Zijlstra wrote: > > On Sun, 2010-02-28 at 20:17 +0100, Raistlin wrote: > > > +/* > > > + * Here we check if --at time t-- a task (which is probably being > > > + * [re]activated or, in general, enqueued) can use its remaining runtime > > > + * and its current deadline _without_ exceeding the bandwidth it is > > > + * assigned (function returns true if it can). > > > + * > > > + * For this to hold, we must check if: > > > + * runtime / (deadline - t) < dl_runtime / dl_deadline . > > > + */ > > > +static bool dl_check_bandwidth(struct sched_dl_entity *dl_se, u64 t) > > > +{ > > > + u64 left, right; > > > + > > > + /* > > > + * left and right are the two sides of the equation above, > > > + * after a bit of shuffling to use multiplications instead > > > + * of divisions. > > > + */ > > > + left = dl_se->dl_deadline * dl_se->runtime; > > > + right = (dl_se->deadline - t) * dl_se->dl_runtime; > > > + > > > + return dl_time_before(left, right); > > > +} > > > > So what happens when we overflow u64? > > Is the resolution in nanosecs starting from zero? If so, then we don't > need to worry about overflow for 583 years? And that is only if the > difference in time is greater than 292 years since dl_time_before() does > a: > > (s64)(a - b) < 0 > > The (s64)(a - b) returns the difference even on overflow as long as the > difference is not greater than 2^63
Its a multiplication of two u64, that's a lot easier to overflow.
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