[PATCH v5 11/17] rbtree: coding style adjustments

[Praveen Kumar <kpraveen.lkml@gmail.com>]

From: Michel Lespinasse <walken@google.com>

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>
[Linux commit 7ce6ff9e5de99e7b72019c7de82fb438fe1dc5a0]

Ported to Xen.

Signed-off-by: Praveen Kumar <kpraveen.lkml@gmail.com>
---
 xen/common/rbtree.c | 42 +++++++++++++++++++++++-------------------
 1 file changed, 23 insertions(+), 19 deletions(-)

diff --git a/xen/common/rbtree.c b/xen/common/rbtree.c
index e58336bd26..215d4f3613 100644
--- a/xen/common/rbtree.c
+++ b/xen/common/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;
@@ -407,17 +406,15 @@ 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);
 }
@@ -459,8 +456,10 @@ 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)
@@ -468,12 +467,13 @@ 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;
 
@@ -488,8 +488,10 @@ 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)
@@ -497,8 +499,10 @@ 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;
 
-- 
2.12.0


_______________________________________________
Xen-devel mailing list
Xen-devel@lists.xen.org
https://lists.xen.org/xen-devel