Re: [PATCH v4 6/8] tracing/histogram: Optimize division by a power of 2

On Tue, Oct 26, 2021 at 5:18 PM Steven Rostedt [off-list ref] wrote:

On Tue, 26 Oct 2021 16:39:13 -0700
Kalesh Singh [off-list ref] wrote:

quoted

quoted

        // 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.

quoted

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

quoted

quoted

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