提交 d33b6fba 编写于 作者: M Matthew Wilcox 提交者: Greg Kroah-Hartman

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: NMatthew Wilcox <matthew@wil.cx>
Cc: Ivan Kokshaysky <ink@jurassic.park.msu.ru>
Cc: Dominik Brodowski <linux@dominikbrodowski.net>
Signed-off-by: NAndrew Morton <akpm@osdl.org>
Signed-off-by: NGreg Kroah-Hartman <gregkh@suse.de>
上级 b2782408
......@@ -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)
{
......@@ -357,20 +356,21 @@ int insert_resource(struct resource *parent, struct resource *new)
struct resource *first, *next;
write_lock(&resource_lock);
begin:
result = 0;
first = __request_resource(parent, new);
if (!first)
goto out;
result = -EBUSY;
if (first == parent)
goto out;
for (;; parent = first) {
result = 0;
first = __request_resource(parent, new);
if (!first)
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;
result = -EBUSY;
if (first == parent)
goto out;
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) {
......
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