On Tue, Oct 26, 2021 at 5:18 PM Steven Rostedt rostedt@goodmis.org wrote:
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.
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