rbtree: coding style adjustments
Set comment and indentation style to be consistent with linux coding style and the rest of the file, as suggested by Peter Zijlstra Signed-off-by: Michel Lespinasse <walken@google.com> Cc: Andrea Arcangeli <aarcange@redhat.com> Acked-by: David Woodhouse <David.Woodhouse@intel.com> Cc: Rik van Riel <riel@redhat.com> Cc: Peter Zijlstra <a.p.zijlstra@chello.nl> Cc: Daniel Santos <daniel.santos@pobox.com> Cc: Jens Axboe <axboe@kernel.dk> Cc: "Eric W. Biederman" <ebiederm@xmission.com> Signed-off-by: Andrew Morton <akpm@linux-foundation.org> Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
This commit is contained in:
Родитель
6280d2356f
Коммит
7ce6ff9e5d
42
lib/rbtree.c
42
lib/rbtree.c
|
@ -363,8 +363,7 @@ void rb_erase(struct rb_node *node, struct rb_root *root)
|
|||
child = node->rb_right;
|
||||
else if (!node->rb_right)
|
||||
child = node->rb_left;
|
||||
else
|
||||
{
|
||||
else {
|
||||
struct rb_node *old = node, *left;
|
||||
|
||||
node = node->rb_right;
|
||||
|
@ -406,17 +405,15 @@ void rb_erase(struct rb_node *node, struct rb_root *root)
|
|||
|
||||
if (child)
|
||||
rb_set_parent(child, parent);
|
||||
if (parent)
|
||||
{
|
||||
if (parent) {
|
||||
if (parent->rb_left == node)
|
||||
parent->rb_left = child;
|
||||
else
|
||||
parent->rb_right = child;
|
||||
}
|
||||
else
|
||||
} else
|
||||
root->rb_node = child;
|
||||
|
||||
color:
|
||||
color:
|
||||
if (color == RB_BLACK)
|
||||
__rb_erase_color(child, parent, root);
|
||||
}
|
||||
|
@ -529,8 +526,10 @@ struct rb_node *rb_next(const struct rb_node *node)
|
|||
if (RB_EMPTY_NODE(node))
|
||||
return NULL;
|
||||
|
||||
/* If we have a right-hand child, go down and then left as far
|
||||
as we can. */
|
||||
/*
|
||||
* If we have a right-hand child, go down and then left as far
|
||||
* as we can.
|
||||
*/
|
||||
if (node->rb_right) {
|
||||
node = node->rb_right;
|
||||
while (node->rb_left)
|
||||
|
@ -538,12 +537,13 @@ struct rb_node *rb_next(const struct rb_node *node)
|
|||
return (struct rb_node *)node;
|
||||
}
|
||||
|
||||
/* No right-hand children. Everything down and left is
|
||||
smaller than us, so any 'next' node must be in the general
|
||||
direction of our parent. Go up the tree; any time the
|
||||
ancestor is a right-hand child of its parent, keep going
|
||||
up. First time it's a left-hand child of its parent, said
|
||||
parent is our 'next' node. */
|
||||
/*
|
||||
* No right-hand children. Everything down and left is smaller than us,
|
||||
* so any 'next' node must be in the general direction of our parent.
|
||||
* Go up the tree; any time the ancestor is a right-hand child of its
|
||||
* parent, keep going up. First time it's a left-hand child of its
|
||||
* parent, said parent is our 'next' node.
|
||||
*/
|
||||
while ((parent = rb_parent(node)) && node == parent->rb_right)
|
||||
node = parent;
|
||||
|
||||
|
@ -558,8 +558,10 @@ struct rb_node *rb_prev(const struct rb_node *node)
|
|||
if (RB_EMPTY_NODE(node))
|
||||
return NULL;
|
||||
|
||||
/* If we have a left-hand child, go down and then right as far
|
||||
as we can. */
|
||||
/*
|
||||
* If we have a left-hand child, go down and then right as far
|
||||
* as we can.
|
||||
*/
|
||||
if (node->rb_left) {
|
||||
node = node->rb_left;
|
||||
while (node->rb_right)
|
||||
|
@ -567,8 +569,10 @@ struct rb_node *rb_prev(const struct rb_node *node)
|
|||
return (struct rb_node *)node;
|
||||
}
|
||||
|
||||
/* No left-hand children. Go up till we find an ancestor which
|
||||
is a right-hand child of its parent */
|
||||
/*
|
||||
* No left-hand children. Go up till we find an ancestor which
|
||||
* is a right-hand child of its parent.
|
||||
*/
|
||||
while ((parent = rb_parent(node)) && node == parent->rb_left)
|
||||
node = parent;
|
||||
|
||||
|
|
Загрузка…
Ссылка в новой задаче