Resources: insert identical resources above existing resources

If you have two resources which aree exactly the same size,
insert_resource() currently inserts the new one below the existing one. 
This is wrong because there's no way to insert a resource of the same size
above an existing one.

I took this opportunity to rewrite the initial loop to be a for-loop
instead of a goto-loop and fix the documentation.
Signed-off-by: Matthew Wilcox <matthew@wil.cx>
Cc: Ivan Kokshaysky <ink@jurassic.park.msu.ru>
Cc: Dominik Brodowski <linux@dominikbrodowski.net>
Signed-off-by: Andrew Morton <akpm@osdl.org>
Signed-off-by: Greg Kroah-Hartman <gregkh@suse.de>

kernel/resource.c

@@ -344,12 +344,11 @@ EXPORT_SYMBOL(allocate_resource);
  *
  * Returns 0 on success, -EBUSY if the resource can't be inserted.
  *
- * This function is equivalent of request_resource when no conflict
+ * This function is equivalent to request_resource when no conflict
  * happens. If a conflict happens, and the conflicting resources
  * entirely fit within the range of the new resource, then the new
- * resource is inserted and the conflicting resources become childs of
- * the new resource.  Otherwise the new resource becomes the child of
- * the conflicting resource
+ * resource is inserted and the conflicting resources become children of
+ * the new resource.
  */
 int insert_resource(struct resource *parent, struct resource *new)
 {
@@ -357,7 +356,8 @@ int insert_resource(struct resource *parent, struct resource *new)
 	struct resource *first, *next;
 
 	write_lock(&resource_lock);
- begin:
+
+	for (;; parent = first) {
 	 	result = 0;
 		first = __request_resource(parent, new);
 		if (!first)
@@ -367,10 +367,10 @@ int insert_resource(struct resource *parent, struct resource *new)
 		if (first == parent)
 			goto out;
 
-	/* Resource fully contained by the clashing resource? Recurse into it */
-	if (first->start <= new->start && first->end >= new->end) {
-		parent = first;
-		goto begin;
+		if ((first->start > new->start) || (first->end < new->end))
+			break;
+		if ((first->start == new->start) && (first->end == new->end))
+			break;
 	}
 
 	for (next = first; ; next = next->sibling) {