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>

diff --git a/lib/rbtree.c b/lib/rbtree.c
index a38e473d8fe7aea4c6c65898b8ff9a85b36da23f..08926709b4f9f16c897c7f38d5290c49419a3c6e 100644 (file)
--- a/lib/rbtree.c
+++ b/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;