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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:
Michel Lespinasse 2012-10-08 16:31:01 -07:00 committed by Linus Torvalds
parent 6280d2356f
commit 7ce6ff9e5d
1 changed files with 23 additions and 19 deletions
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42
lib/rbtree.c
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@ -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;