linux-kernel - Re: [RFC PATCH] clk: fractional-divider: Correct max_{m,n} handed over to rational_best

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Date:   Wed, 14 Jul 2021 13:38:22 +0300
From:   Andy Shevchenko <andriy.shevchenko@...ux.intel.com>
To:     Liu Ying <victor.liu@....com>
Cc:     linux-clk@...r.kernel.org, linux-kernel@...r.kernel.org,
        Heikki Krogerus <heikki.krogerus@...ux.intel.com>,
        Michael Turquette <mturquette@...libre.com>,
        Stephen Boyd <sboyd@...nel.org>,
        Dong Aisheng <aisheng.dong@....com>,
        NXP Linux Team <linux-imx@....com>,
        Jacky Bai <ping.bai@....com>
Subject: Re: [RFC PATCH] clk: fractional-divider: Correct max_{m,n} handed
 over to rational_best_approximation()

On Wed, Jul 14, 2021 at 06:10:46PM +0800, Liu Ying wrote:
> On Wed, 2021-07-14 at 12:12 +0300, Andy Shevchenko wrote:
> > On Wed, Jul 14, 2021 at 02:41:29PM +0800, Liu Ying wrote:

...

> > > /*
> > >  * Get rate closer to *parent_rate to guarantee there is no overflow
> > >  * for m and n. In the result it will be the nearest rate left shifted
> > >  * by (scale - fd->nwidth) bits.
> > >  */
> > 
> > I don't know how to rephrase above comment better.
> > 
> > > scale = fls_long(*parent_rate / rate - 1);
> > > if (scale > fd->nwidth)
> > > 	rate <<= scale - fd->nwidth;
> > 
> > This takes an advantage of the numbers be in a form of
> > 
> > 	n = k * 2^m, (1)
> > 
> > where m will be scale in the snippet above. Thus, if n can be represented by
> > (1), we opportunistically reduce amount of bits needed for it by shifting right
> > by m bits.
> > 
> > Does it make sense?
> 
> Thanks for your explaination.
> But, sorry, Jacky and I still don't understand this.

Okay, We have two values in question:
 r_o (original rate of the parent clock)
 r_u (the rate user asked for)

We have a pre-scaler block which asks for
 m (denominator)
 n (nominator)
values to be provided to satisfy the (2)

	r_u ~= r_o * m / n, (2)

where we try our best to make it "=" instead of "~=".

Now, m and n have the limitation by a range, e.g.

	n >= 1, n < N_lim, where N_lim = 2^nlim. (3)

Hence, from (2) and (3), assuming the worst case m = 1,

	ln2(r_o / r_u) <= nlim. (4)

The above code tries to satisfy (4).

Have you got it now?

> > The code looks good to me, btw, although I dunno if you need to call the newly
> > introduced function before or after the above mentioned snippet.
> 
> Assuming that snippet is fully orthogonal to this patch, then it
> doesn't matter if it's before or after.

Please, double check this. Because you play with limits, while we expect them
to satisfy (3).

-- 
With Best Regards,
Andy Shevchenko