On Sun, Oct 26, 2008 at 05:48:47PM +0100, Ilko Iliev wrote:
With this patch "nand erase clean" writes correctly the cleanmarkers. Without this patch "nand erase clean" fills the OOB with zeros which marks all blocks as bad.
Signed-off-by: Ilko Iliev iliev@ronetix.at
drivers/mtd/nand/nand_util.c | 27 +++++++++++++++++++-------- 1 files changed, 19 insertions(+), 8 deletions(-)
diff --git a/drivers/mtd/nand/nand_util.c b/drivers/mtd/nand/nand_util.c index 52b3d21..a601772 100644 --- a/drivers/mtd/nand/nand_util.c +++ b/drivers/mtd/nand/nand_util.c @@ -156,10 +156,19 @@ int nand_erase_opts(nand_info_t *meminfo, const nand_erase_options_t *opts) /* format for JFFS2 ? */ if (opts->jffs2) {
chip->ops.len = chip->ops.ooblen = 64;
if ( chip->ecc.layout->oobfree->length <cleanmarker.totlen ) {
Patch is linewrapped. Also, no space after ( or before ).
Why must the cleanmarker fit in the first free segment?
memset(buf, 0xFF, sizeof(buf));chip->ops.oobbuf = buf;chip->ops.ooboffs = chip->badblockpos &~0x01;
chip->ops.len = chip->ops.ooblen =meminfo->oobsize;
What if oobsize > 64 (as with 4k pages)? Why write anything at all if you're not going to write the cleanmarker? Why badblockpos & ~1 (I know existing code does it, but why)?
}else {
} else {
chip->ops.oobbuf = (uint8_t *)&cleanmarker;chip->ops.ooboffs =chip->ecc.layout->oobfree->offset;
chip->ops.len = chip->ops.ooblen =cleanmarker.totlen;
}
Set ooboffs to zero, and use MTD_OOB_AUTO.
-Scott