Messages in this thread | | | Date | Tue, 15 May 2018 22:35:16 +0200 | From | Boris Brezillon <> | Subject | Re: [PATCH v5 2/2] mtd: rawnand: use bit-wise majority to recover the contents of ONFI parameter |
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On Tue, 15 May 2018 23:23:02 +0300 Andy Shevchenko <andy.shevchenko@gmail.com> wrote:
> 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))
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. >
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