in Documentation/filesystems/propagate_umount.txt:
line 289: remove whitespace on blank line
line 315: remove duplicate "that"
line 364: remove duplicate "in"
Signed-off-by: Raphael Pinsonneault-Thibeault <rpthibeault@xxxxxxxxx>
---
Documentation/filesystems/propagate_umount.txt | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/Documentation/filesystems/propagate_umount.txt b/Documentation/filesystems/propagate_umount.txt
index c90349e5b889..9a7eb96df300 100644
--- a/Documentation/filesystems/propagate_umount.txt
+++ b/Documentation/filesystems/propagate_umount.txt
@@ -286,7 +286,7 @@ Trim_one(m)
strip the "seen by Trim_ancestors" mark from m
remove m from the Candidates list
return
-
+
remove_this = false
found = false
for each n in children(m)
@@ -312,7 +312,7 @@ Trim_ancestors(m)
}
Terminating condition in the loop in Trim_ancestors() is correct,
-since that that loop will never run into p belonging to U - p is always
+since that loop will never run into p belonging to U - p is always
an ancestor of argument of Trim_one() and since U is closed, the argument
of Trim_one() would also have to belong to U. But Trim_one() is never
called for elements of U. In other words, p belongs to S if and only
@@ -361,7 +361,7 @@ such removals.
Proof: suppose S was non-shifting, x is a locked element of S, parent of x
is not in S and S - {x} is not non-shifting. Then there is an element m
in S - {x} and a subtree mounted strictly inside m, such that m contains
-an element not in in S - {x}. Since S is non-shifting, everything in
+an element not in S - {x}. Since S is non-shifting, everything in
that subtree must belong to S. But that means that this subtree must
contain x somewhere *and* that parent of x either belongs that subtree
or is equal to m. Either way it must belong to S. Contradiction.
--
2.43.0
