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....@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?

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

Reply via email to