`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 2^63Its a multiplication of two u64, that's a lot easier to overflow.`