Hi,
> > >
> > > ser_clk = 0;
> > > maxdiv = CLK_DIV_MSK >> CLK_DIV_SHFT;
> > > div = 1;
> > > while (div < maxdiv) {
> >
> >
> > div <= maxdiv ?
>
> Ah, sure.
Thank you.
>
>
> > > mult = (unsigned long long)div * desired_clk;
> > > if (mult != (unsigned long)mult)
> > > break;
> > >
> > > two_percent = mult / 50;
> > >
> > > /*
> > > * Loop requesting (freq - 2%) and possibly (freq).
> > > *
> > > * We'll keep track of the lowest freq inexact match we found
> > > * but always try to find a perfect match. NOTE: this algorithm
> > > * could miss a slightly better freq if there's more than one
> > > * freq between (freq - 2%) and (freq) but (freq) can't be made
> > > * exactly, but that's OK.
> > > *
> > > * This absolutely relies on the fact that the Qualcomm clock
> > > * driver always rounds up.
> > > */
> > > test_freq = mult - two_percent;
> > > while (test_freq <= mult) {
> > > freq = clk_round_rate(clk, test_freq);
> > >
> > > /*
> > > * A dead-on freq is an insta-win. This implicitly
> > > * handles when "freq == mult"
> > > */
> > > if (!(freq % desired_clk)) {
> > > *clk_div = freq / desired_clk;
> > > return freq;
> > > }
> > >
> > > /*
> > > * Only time clock framework doesn't round up is if
> > > * we're past the max clock rate. We're done searching
> > > * if that's the case.
> > > */
> > > if (freq < test_freq)
> > > return ser_clk;
> > >
> > > /* Save the first (lowest freq) within 2% */
> > > if (!ser_clk && freq <= mult + two_percent) {
> > > ser_clk = freq;
> > > *clk_div = div;
> > > }
> >
> > My last concern is with search happening only within 2% tolerance.
> > Do we fail otherwise?
> >
> > This real case has best tolerance of 1.9% and seems close.
> >
> > [ 17.963672] 20220530 desired_clk-51200000
> > [ 21.193550] 20220530 returning ser_clk-52174000, div-1, diff-974000
> >
> > Perhaps we can fallback on 1st clock rate?
>
> I don't feel super comfortable just blindly falling back on the 1st clock rate. It
> could be wildly (more than 5%) wrong, can't it?
>
> IMO:
> * If you're not comfortable with 2%, you could always pick 3% or 4%.
> As I said, my random web search seemed to indicate that up to 5% was
> perhaps OK.
> * It's probably overkill, but you could abstract the whole search out and try
> searching once for 2% and then try 4%?
>
Ok, I will implement a function that searches within an input tolerance.
And have a conditional 2nd call to same with higher tolerance of 5%.
This would mean that we will still run through 2 loops in some cases, but that’s ok.
Thank you.
>
> -Doug
