Messages in this thread | | | Date | Tue, 26 Oct 2021 20:18:46 -0400 | From | Steven Rostedt <> | Subject | Re: [PATCH v4 6/8] tracing/histogram: Optimize division by a power of 2 |
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On Tue, 26 Oct 2021 16:39:13 -0700 Kalesh Singh <kaleshsingh@google.com> 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
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