diff options
| author | Shinya Kuribayashi <shinya.kuribayashi.px@renesas.com> | 2010-05-06 19:22:09 +0900 |
|---|---|---|
| committer | Artem Bityutskiy <Artem.Bityutskiy@nokia.com> | 2010-05-07 08:33:32 +0300 |
| commit | be436f6238a17b8432b9de0212bcfc838afb1f85 (patch) | |
| tree | 59b685247d6d3647cc82920092759b4b09599aa5 /drivers/mtd | |
| parent | 3f5026222e8a16daaa830eec4d72c6745b74407e (diff) | |
UBI: misc comment fixes
Signed-off-by: Shinya Kuribayashi <shinya.kuribayashi.px@renesas.com>
Signed-off-by: Artem Bityutskiy <Artem.Bityutskiy@nokia.com>
Diffstat (limited to 'drivers/mtd')
| -rw-r--r-- | drivers/mtd/ubi/io.c | 6 | ||||
| -rw-r--r-- | drivers/mtd/ubi/vtbl.c | 4 |
2 files changed, 5 insertions, 5 deletions
diff --git a/drivers/mtd/ubi/io.c b/drivers/mtd/ubi/io.c index 016ec1387bc2..4b979e34b159 100644 --- a/drivers/mtd/ubi/io.c +++ b/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/drivers/mtd/ubi/vtbl.c b/drivers/mtd/ubi/vtbl.c index cd90ff3b76b1..14c10bed94ee 100644 --- a/drivers/mtd/ubi/vtbl.c +++ b/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); |