[PATCH] drivers/base/node: Fix double free in register_one_node()

[Donet Tom <donettom@linux.ibm.com>]

When device_register() fails in register_node(), it calls
put_device(&node->dev). This triggers node_device_release(),
which calls kfree(to_node(dev)), thereby freeing the entire
node structure.

As a result, when register_node() returns an error, the node
memory has already been freed. Calling kfree(node) again in
register_one_node() leads to a double free.

This patch removes the redundant kfree(node) from
register_one_node() to prevent the double free.

Fixes: 786eb990cfb7 ("drivers/base/node: handle error properly in register_one_node()")
Signed-off-by: Donet Tom <donettom@linux.ibm.com>
---
 drivers/base/node.c | 1 -
 1 file changed, 1 deletion(-)

diff --git a/drivers/base/node.c b/drivers/base/node.c
index 1608816de67f..6b6e55a98b79 100644
--- a/drivers/base/node.c
+++ b/drivers/base/node.c
@@ -885,7 +885,6 @@ int register_one_node(int nid)
 	error = register_node(node_devices[nid], nid);
 	if (error) {
 		node_devices[nid] = NULL;
-		kfree(node);
 		return error;
 	}
 
-- 
2.51.0