Re: [PATCH v7] mtd: rawnand: use bit-wise majority to recover the contents of ONFI parameter

Boris Brezillon
2018-05-15T07:34:58Z

On Tue, 15 May 2018 06:45:51 +0200
Chris Moore <moore@free.fr> wrote:

> Hi,
>
> Le 13/05/2018 à 06:30, Wan, Jane (Nokia - US/Sunnyvale) a écrit :
> > Per ONFI specification (Rev. 4.0), if all parameter pages have invalid CRC values, the bit-wise majority may be used to recover the contents of the parameter pages from the parameter page copies present.
> >
> > Signed-off-by: Jane Wan <Jane.Wan@nokia.com>
> > ---
> > v7: change debug print messages
> > v6: support the cases that srcbufs are not contiguous
> > v5: make the bit-wise majority functon generic
> > v4: move the bit-wise majority code in a separate function
> > v3: fix warning message detected by kbuild test robot
> > v2: rebase the changes on top of v4.17-rc1
> >
> > drivers/mtd/nand/raw/nand_base.c | 50 ++++++++++++++++++++++++++++++++++----
> > 1 file changed, 45 insertions(+), 5 deletions(-)
> >
> > diff --git a/drivers/mtd/nand/raw/nand_base.c b/drivers/mtd/nand/raw/nand_base.c
> > index 72f3a89..b43b784 100644
> > --- a/drivers/mtd/nand/raw/nand_base.c
> > +++ b/drivers/mtd/nand/raw/nand_base.c
> > @@ -5087,6 +5087,35 @@ static int nand_flash_detect_ext_param_page(struct nand_chip *chip,
> > }
> >
> > /*
> > + * Recover data with bit-wise majority
> > + */
> > +static void nand_bit_wise_majority(const void **srcbufs,
> > + unsigned int nsrcbufs,
> > + void *dstbuf,
> > + unsigned int bufsize)
> > +{
> > + int i, j, k;
> > +
> > + for (i = 0; i < bufsize; i++) {
> > + u8 cnt, val;
> > +
> > + val = 0;
> > + for (j = 0; j < 8; j++) {
> > + cnt = 0;
> > + for (k = 0; k < nsrcbufs; k++) {
> > + const u8 *srcbuf = srcbufs[k];
> > +
> > + if (srcbuf[i] & BIT(j))
> > + cnt++;
> > + }
> > + if (cnt > nsrcbufs / 2)
> > + val |= BIT(j);
> > + }
> > + ((u8 *)dstbuf)[i] = val;
> > + }
> > +}
> > +
> > +/*
> > * Check if the NAND chip is ONFI compliant, returns 1 if it is, 0 otherwise.
> > */
> > static int nand_flash_detect_onfi(struct nand_chip *chip)
> > @@ -5102,7 +5131,7 @@ static int nand_flash_detect_onfi(struct nand_chip *chip)
> > return 0;
> >
> > /* ONFI chip: allocate a buffer to hold its parameter page */
> > - p = kzalloc(sizeof(*p), GFP_KERNEL);
> > + p = kzalloc((sizeof(*p) * 3), GFP_KERNEL);
> > if (!p)
> > return -ENOMEM;
> >
> > @@ -5113,21 +5142,32 @@ static int nand_flash_detect_onfi(struct nand_chip *chip)
> > }
> >
> > for (i = 0; i < 3; i++) {
> > - ret = nand_read_data_op(chip, p, sizeof(*p), true);
> > + ret = nand_read_data_op(chip, &p[i], sizeof(*p), true);
> > if (ret) {
> > ret = 0;
> > goto free_onfi_param_page;
> > }
> >
> > - if (onfi_crc16(ONFI_CRC_BASE, (uint8_t *)p, 254) ==
> > + if (onfi_crc16(ONFI_CRC_BASE, (u8 *)&p[i], 254) ==
> > le16_to_cpu(p->crc)) {
> > + if (i)
> > + memcpy(p, &p[i], sizeof(*p));
> > break;
> > }
> > }
> >
> > if (i == 3) {
> > - pr_err("Could not find valid ONFI parameter page; aborting\n");
> > - goto free_onfi_param_page;
> > + const void *srcbufs[3] = {p, p + 1, p + 2};
> > +
> > + pr_warn("Could not find a valid ONFI parameter page, trying bit-wise majority to recover it\n");
> > + nand_bit_wise_majority(srcbufs, ARRAY_SIZE(srcbufs), p,
> > + sizeof(*p));
> > +
> > + if (onfi_crc16(ONFI_CRC_BASE, (u8 *)p, 254) !=
> > + le16_to_cpu(p->crc)) {
> > + pr_err("ONFI parameter recovery failed, aborting\n");
> > + goto free_onfi_param_page;
> > + }
> > }
> >
> > /* Check version */
>
> This version is still hard coded for a three sample bitwise majority vote.
> So why not use the method which I suggested previously for v2 and which
> I repeat below?

Because I want the nand_bit_wise_majority() function to work with
nsrcbufs > 3 (the ONFI spec says there's at least 3 copy of the param
page, but NAND vendor can decide to put more). Also, if the X copies of
the PARAM are corrupted (which is rather unlikely), that means we
already spent quite a lot of time reading the different copies and
calculating the CRC, so I think we don't care about perf optimizations
when doing bit-wise majority.

>
> The three sample bitwise majority can be implemented without bit level
> manipulation using the identity:
> majority3(a, b, c) = (a & b) | (a & c) | (b & c)
> This can be factorized slightly to (a & (b | c)) | (b & c)
> This enables the operation to be performed 8, 16, 32 or even 64 bits at
> a time depending on the hardware.
>
> This method is not only faster and but also more compact.
>
> Cheers,
> Chris
>
>
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