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authorAdrian Hunter <ext-adrian.hunter@nokia.com>2008-07-30 12:18:02 +0300
committerArtem Bityutskiy <Artem.Bityutskiy@nokia.com>2008-08-13 11:39:20 +0300
commit3a13252c6f3a029ac992a36910e945f361482797 (patch)
treec9d62f776c9761bd15d9b1b5c05f032895c7e8a9
parent22bc7fa8c5da09805edc6a6199ce81373b2c207d (diff)
UBIFS: correct spelling of "thrice".
Signed-off-by: Adrian Hunter <ext-adrian.hunter@nokia.com>
-rw-r--r--fs/ubifs/budget.c4
-rw-r--r--fs/ubifs/find.c2
2 files changed, 3 insertions, 3 deletions
diff --git a/fs/ubifs/budget.c b/fs/ubifs/budget.c
index 323d83a4d099..154098157473 100644
--- a/fs/ubifs/budget.c
+++ b/fs/ubifs/budget.c
@@ -263,7 +263,7 @@ int ubifs_calc_min_idx_lebs(struct ubifs_info *c)
idx_size = c->old_idx_sz + c->budg_idx_growth + c->budg_uncommitted_idx;
- /* And make sure we have trice the index size of space reserved */
+ /* And make sure we have thrice the index size of space reserved */
idx_size = idx_size + (idx_size << 1);
/*
@@ -388,7 +388,7 @@ static int can_use_rp(struct ubifs_info *c)
* This function makes sure UBIFS has enough free eraseblocks for index growth
* and data.
*
- * When budgeting index space, UBIFS reserves trice as more LEBs as the index
+ * When budgeting index space, UBIFS reserves thrice as many LEBs as the index
* would take if it was consolidated and written to the flash. This guarantees
* that the "in-the-gaps" commit method always succeeds and UBIFS will always
* be able to commit dirty index. So this function basically adds amount of
diff --git a/fs/ubifs/find.c b/fs/ubifs/find.c
index c70c7679c1bf..adee7b5ddeab 100644
--- a/fs/ubifs/find.c
+++ b/fs/ubifs/find.c
@@ -290,7 +290,7 @@ int ubifs_find_dirty_leb(struct ubifs_info *c, struct ubifs_lprops *ret_lp,
idx_lp = idx_heap->arr[0];
sum = idx_lp->free + idx_lp->dirty;
/*
- * Since we reserve trice as more space for the index than it
+ * Since we reserve thrice as much space for the index than it
* actually takes, it does not make sense to pick indexing LEBs
* with less than, say, half LEB of dirty space. May be half is
* not the optimal boundary - this should be tested and