On Tue, Oct 26, 2021 at 6:15 PM Steven Rostedt [off-list ref] wrote:

On Tue, 26 Oct 2021 18:09:22 -0700
Kalesh Singh [off-list ref] wrote:

quoted

quoted

  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:

That should have been (as shown in the algorithm)

  delta = mult * div - 2 ^ 20

As mult is 2^20 / div; and the above should end up zero if there's no
rounding issues, as it would be:

 delta = (2^20 / div) * div - 2^20

Good catch. We're checking if we get back the exact value.

And IIUC max_div is an arbitrary value we decide on that's <= 2^shift?
Is there a rule of thumb for choosing this?

Thanks,
Kalesh

-- Steve