Re: [PATCH] mtd: nand: gpmi: Fix ecc strength calculation

Han Xu
2016-06-21T15:52:33Z

On Tue, Jun 21, 2016 at 9:46 AM, Boris Brezillon
<boris.brezillon@free-electrons.com> wrote:
> On Tue, 21 Jun 2016 16:35:49 +0200
> Sascha Hauer <s.hauer@pengutronix.de> wrote:
>
>> BCH ECC correction works in chunks of 512 bytes, so a 2k page size nand
>> is divided into 4 chunks. Hardware requires that each chunk has a full
>> number of bytes, so when we need 9 bits per chunk we must round up to
>> two bytes. The current code misses that and calculates a ECC strength
>> of 18 for a 2048+128 byte page size NAND. ECC strength of 18 requires
>> 30 bytes per chunk, so a total of 4 * (512 + 30) + 10 = 2178 bytes when
>> the device only has a page size of 2176 bytes.
>
> AFAIR, the GPMI/ECC engine operates at the bit level (which is a pain
> to deal with BTW), and is only requiring a byte alignment on the total
> number of ECC bits. So here, DIV_ROUND_UP(18 * 13 * 4, 8) = 117, which
> fits in the 118 bytes (128 bytes - 10 bytes of 'metadata').
>
> Han, can you confirm that?

Correct, BCH module works at bit level, 18bit ECC won't exceed the oob size.

>
>>
>> Fix this by first calculating the number of bytes per chunk we have
>> available for ECC which also makes it easier to follow the calculation.
>>
>> Signed-off-by: Sascha Hauer <s.hauer@pengutronix.de>
>> ---
>> drivers/mtd/nand/gpmi-nand/gpmi-nand.c | 28 ++++++++--------------------
>> 1 file changed, 8 insertions(+), 20 deletions(-)
>>
>> diff --git a/drivers/mtd/nand/gpmi-nand/gpmi-nand.c b/drivers/mtd/nand/gpmi-nand/gpmi-nand.c
>> index 6e46156..95fb3dc 100644
>> --- a/drivers/mtd/nand/gpmi-nand/gpmi-nand.c
>> +++ b/drivers/mtd/nand/gpmi-nand/gpmi-nand.c
>> @@ -119,32 +119,20 @@ static irqreturn_t bch_irq(int irq, void *cookie)
>> return IRQ_HANDLED;
>> }
>>
>> -/*
>> - * Calculate the ECC strength by hand:
>> - * E : The ECC strength.
>> - * G : the length of Galois Field.
>> - * N : The chunk count of per page.
>> - * O : the oobsize of the NAND chip.
>> - * M : the metasize of per page.
>> - *
>> - * The formula is :
>> - * E * G * N
>> - * ------------ <= (O - M)
>> - * 8
>> - *
>> - * So, we get E by:
>> - * (O - M) * 8
>> - * E <= -------------
>> - * G * N
>> - */
>> static inline int get_ecc_strength(struct gpmi_nand_data *this)
>> {
>> struct bch_geometry *geo = &this->bch_geometry;
>> struct mtd_info *mtd = nand_to_mtd(&this->nand);
>> int ecc_strength;
>> + int n;
>> +
>> + /* number of ecc bytes we have per chunk */
>> + n = (mtd->oobsize - geo->metadata_size) / geo->ecc_chunk_count;
>> +
>> + /* in bits */
>> + n <<= 3;
>>
>> - ecc_strength = ((mtd->oobsize - geo->metadata_size) * 8)
>> - / (geo->gf_len * geo->ecc_chunk_count);
>> + ecc_strength = n / geo->gf_len;
>>
>> /* We need the minor even number. */
>> return round_down(ecc_strength, 2);
>
>
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--
Sincerely,

Han XU