of: fix of_node leak caused in of_find_node_opts_by_path

During stepping down the tree, parent node is gotten first and its refcount is
increased with of_node_get() in __of_get_next_child(). Since it just being used
as tmp node, its refcount must be decreased with of_node_put() after traversing
its child nodes.

Or, its refcount will never be descreased to ZERO, then it will never be freed,
as well as other related memory blocks.

To fix this, decrease refcount of parent with of_node_put() after
__of_find_node_by_path().

Signed-off-by: Qi Hou <qi.hou@windriver.com>
Acked-by: Peter Rosin <peda@axentia.se>
Signed-off-by: Rob Herring <robh@kernel.org>
This commit is contained in:
Qi Hou 2017-02-06 12:55:19 +08:00 committed by Rob Herring
parent 4b741bc359
commit 0549bde0fc
1 changed files with 3 additions and 0 deletions

View File

@ -842,8 +842,11 @@ struct device_node *of_find_node_opts_by_path(const char *path, const char **opt
if (!np) if (!np)
np = of_node_get(of_root); np = of_node_get(of_root);
while (np && *path == '/') { while (np && *path == '/') {
struct device_node *tmp = np;
path++; /* Increment past '/' delimiter */ path++; /* Increment past '/' delimiter */
np = __of_find_node_by_path(np, path); np = __of_find_node_by_path(np, path);
of_node_put(tmp);
path = strchrnul(path, '/'); path = strchrnul(path, '/');
if (separator && separator < path) if (separator && separator < path)
break; break;