[<prev] [next>] [<thread-prev] [thread-next>] [day] [month] [year] [list]
Message-ID: <20251031124530.3db7805b@pumpkin>
Date: Fri, 31 Oct 2025 12:45:30 +0000
From: David Laight <david.laight.linux@...il.com>
To: Andy Shevchenko <andriy.shevchenko@...ux.intel.com>
Cc: Jonathan Cameron <Jonathan.Cameron@...wei.com>,
linux-iio@...r.kernel.org, linux-kernel@...r.kernel.org, Jyoti Bhayana
<jbhayana@...gle.com>, Jonathan Cameron <jic23@...nel.org>, David Lechner
<dlechner@...libre.com>, Nuno Sá <nuno.sa@...log.com>, Andy
Shevchenko <andy@...nel.org>
Subject: Re: [PATCH v1 1/1] iio: common: scmi_sensors: Replace const_ilog2()
with ilog2()
On Fri, 31 Oct 2025 11:54:30 +0200
Andy Shevchenko <andriy.shevchenko@...ux.intel.com> wrote:
> On Fri, Oct 31, 2025 at 09:43:36AM +0000, David Laight wrote:
> > On Fri, 31 Oct 2025 08:45:00 +0100
> > Andy Shevchenko <andriy.shevchenko@...ux.intel.com> wrote:
>
> ...
>
> > > tstamp_scale = sensor->sensor_info->tstamp_scale +
> > > - const_ilog2(NSEC_PER_SEC) / const_ilog2(10);
> > > + ilog2(NSEC_PER_SEC) / ilog2(10);
> >
> > Is that just a strange way of writing 9 ?
>
> Why? It's correct way of writing log¹⁰(NSEC_PER_SEC), the problem here is that
> "i" people do not think about :-)
Even without the "i" the division could easily give 8.999999.
So you'd be relying on rounding to get the required integral value.
> But we have intlog10(), I completely forgot about it.
And it isn't the function the code is looking for.
(The result is shifted left 24 and it doesn't have an optimisation
for constants.)
>
> > Mathematically log2(x)/log2(10) is log10(x) - which would be 9.
> > The code does seem to be 'in luck' though.
> > NSEC_PER_SEC is 10^9 or 0x3b9aca00, so ilog2(NSEC_PER_SEC) is 29.
> > ilog2(10) is 3, and 29/3 is 9.
> >
> > Do the same for 10^10 and you get 11.
>
> That code looks like working by luck entirely, TBH. I just took the scope of
> the patch to start dropping const_ilog2() usages.
Something always crawls out of the woodwork...
David
Powered by blists - more mailing lists