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.
> >
> >
> > 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
