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From:   Andy Shevchenko <andy.shevchenko@gmail.com>
Date:   Tue, 15 May 2018 23:23:02 +0300
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Subject: Re: [PATCH v5 2/2] mtd: rawnand: use bit-wise majority to recover the
 contents of ONFI parameter
To:     Boris Brezillon <boris.brezillon@bootlin.com>
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        Masahiro Yamada <yamada.masahiro@socionext.com>,
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On Tue, May 15, 2018 at 11:03 AM, Boris Brezillon
<boris.brezillon@bootlin.com> wrote:
> On Tue, 15 May 2018 10:46:00 +0300
> Andy Shevchenko <andy.shevchenko@gmail.com> wrote:
>
>> On Tue, May 15, 2018 at 10:35 AM, Boris Brezillon
>> <boris.brezillon@bootlin.com> wrote:
>> > On Mon, 14 May 2018 20:54:36 +0300
>> > Andy Shevchenko <andy.shevchenko@gmail.com> wrote:

>> >> >                         for (k = 0; k < nbufs; k++) {
>> >> >                                 const u8 *srcbuf = srcbufs[j];
>> >> >
>> >> >                                 if (srcbuf[i] & BIT(k))
>> >> >                                         m++;
>> >> >                         }
>> >>
>> >> ...which is effectively hweightXX().
>> >
>> > No it's not.
>>
>> I don't see how "not". In the loop everithing except m and k are
>> invariants. What did I miss?
>
> We're not counting the number of bits set in an uXX var, but the number
> of set bits at the same position in different buffers.

...on big picture. The excerpt above is hweight() against srcbuf[i].

Let's rewrite it like this:

                    const u8 *srcbuf = srcbufs[j];

                    for (k = 0; k < nbufs; k++) {
                               if (srcbuf[i] & BIT(k))
                                      m++;
                    }

...and now it looks obvious:

m += hweight...(srcbuf[i])

_If_ nbufs is power of two we may use primitive helper.

-- 
With Best Regards,
Andy Shevchenko