@@ -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,
@@ -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)
- * 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);