On Tue, Oct 26, 2021 at 5:18 PM Steven Rostedt <rostedt@xxxxxxxxxxx> wrote:
>
> On Tue, 26 Oct 2021 16:39:13 -0700
> Kalesh Singh <kaleshsingh@xxxxxxxxxx> wrote:
>
> > > // This works best for small divisors
> > > if (div > max_div) {
> > > // only do a real division
> > > return;
> > > }
> > > shift = 20;
> > > mult = ((1 << shift) + div - 1) / div;
> > > delta = mult * div - (1 << shift);
> > > if (!delta) {
> > > /* div is a power of 2 */
> > > max = -1;
> > > return;
> > > }
> > > max = (1 << shift) / delta;
> >
> > I'm still trying to digest the above algorithm.
>
> mult = (2^20 + div - 1) / div;
>
> The "div - 1" is to round up.
>
> Basically, it's doing: X / div = X * (2^20 / div) / 2^20
>
> If div is constant, the 2^20 / div is constant, and the "2^20" is the
> same as a shift.
>
> So multiplier is 2^20 / div, and the shift is 20.
>
> But because there's rounding errors it is only accurate up to the
> difference of:
>
> delta = mult * div / 2^20
>
> That is if mult is a power of two, then there would be no rounding
> errors, and the delta is zero, making the max infinite:
>
> max = 2^20 / delta as delta goes to zero.
>
> > But doesn't this add 2 extra divisions? What am I missing here?
>
> The above is only done at parsing not during the trace, where we care
> about.
Hi Steve,
Thanks for the explanation, this cleared it up for me.
- Kalesh
>
> > >
> > >
> > > We would of course need to use 64 bit operations (maybe only do this for 64
> > > bit machines). And perhaps even use bigger shift values to get a bigger max.
> > >
> > > Then we could do:
> > >
> > > if (val1 < max)
> > > return (val1 * mult) >> shift;
>
> This is done at the time of recording.
>
> Actually, it would be:
>
> if (val1 < max)
> return (val1 * mult) >> shift;
> else
> return val1 / div;
>
> -- Steve
