On Tue, 2010-04-13 at 14:55 -0400, Steven Rostedt wrote:
quoted text
> 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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