[PATCH v5 2/2] mtd: rawnand: use bit-wise majority to recover the contents of ONFI parameter

[Boris Brezillon]

On Tue, 15 May 2018 23:23:02 +0300
Andy Shevchenko <andy.shevchenko at gmail.com> wrote:

> On Tue, May 15, 2018 at 11:03 AM, Boris Brezillon
> <boris.brezillon at bootlin.com> wrote:
> > On Tue, 15 May 2018 10:46:00 +0300
> > Andy Shevchenko <andy.shevchenko at gmail.com> wrote:
> >  
> >> On Tue, May 15, 2018 at 10:35 AM, Boris Brezillon
> >> <boris.brezillon at bootlin.com> wrote:  
> >> > On Mon, 14 May 2018 20:54:36 +0300
> >> > Andy Shevchenko <andy.shevchenko at 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))

I made a mistake in my code sample, it's

				if (srcbuf[i] & BIT(j))

If you look at v6, you'll see it's been fixed by Jane.

>                                       m++;
>                     }
> 
> ...and now it looks obvious:
> 
> m += hweight...(srcbuf[i])
> 
> _If_ nbufs is power of two we may use primitive helper.
>