From 56af58dc29b0d742cc8bef8ac749d340f67796ac Mon Sep 17 00:00:00 2001 From: Shinya Kuribayashi Date: Thu, 6 May 2010 19:22:09 +0900 Subject: [PATCH] --- yaml --- r: 195832 b: refs/heads/master c: be436f6238a17b8432b9de0212bcfc838afb1f85 h: refs/heads/master v: v3 --- [refs] | 2 +- trunk/drivers/mtd/ubi/io.c | 6 +++--- trunk/drivers/mtd/ubi/vtbl.c | 4 ++-- 3 files changed, 6 insertions(+), 6 deletions(-) diff --git a/[refs] b/[refs] index ab099531dc1b..fa18bd0f333e 100644 --- a/[refs] +++ b/[refs] @@ -1,2 +1,2 @@ --- -refs/heads/master: 3f5026222e8a16daaa830eec4d72c6745b74407e +refs/heads/master: be436f6238a17b8432b9de0212bcfc838afb1f85 diff --git a/trunk/drivers/mtd/ubi/io.c b/trunk/drivers/mtd/ubi/io.c index 016ec1387bc2..4b979e34b159 100644 --- a/trunk/drivers/mtd/ubi/io.c +++ b/trunk/drivers/mtd/ubi/io.c @@ -64,9 +64,9 @@ * device, e.g., make @ubi->min_io_size = 512 in the example above? * * A: because when writing a sub-page, MTD still writes a full 2K page but the - * bytes which are no relevant to the sub-page are 0xFF. So, basically, writing - * 4x512 sub-pages is 4 times slower than writing one 2KiB NAND page. Thus, we - * prefer to use sub-pages only for EV and VID headers. + * bytes which are not relevant to the sub-page are 0xFF. So, basically, + * writing 4x512 sub-pages is 4 times slower than writing one 2KiB NAND page. + * Thus, we prefer to use sub-pages only for EC and VID headers. * * As it was noted above, the VID header may start at a non-aligned offset. * For example, in case of a 2KiB page NAND flash with a 512 bytes sub-page, diff --git a/trunk/drivers/mtd/ubi/vtbl.c b/trunk/drivers/mtd/ubi/vtbl.c index cd90ff3b76b1..14c10bed94ee 100644 --- a/trunk/drivers/mtd/ubi/vtbl.c +++ b/trunk/drivers/mtd/ubi/vtbl.c @@ -414,7 +414,7 @@ static struct ubi_vtbl_record *process_lvol(struct ubi_device *ubi, * 0 contains more recent information. * * So the plan is to first check LEB 0. Then - * a. if LEB 0 is OK, it must be containing the most resent data; then + * a. if LEB 0 is OK, it must be containing the most recent data; then * we compare it with LEB 1, and if they are different, we copy LEB * 0 to LEB 1; * b. if LEB 0 is corrupted, but LEB 1 has to be OK, and we copy LEB 1 @@ -848,7 +848,7 @@ int ubi_read_volume_table(struct ubi_device *ubi, struct ubi_scan_info *si) goto out_free; /* - * Get sure that the scanning information is consistent to the + * Make sure that the scanning information is consistent to the * information stored in the volume table. */ err = check_scanning_info(ubi, si);