Re: [PATCH v2] drivers/base/node: Handle error properly in register_one_node()

From: Donet Tom

Date: Wed Sep 17 2025 - 11:03:57 EST



On 9/17/25 7:15 PM, Chris Mason wrote:
On Fri, 22 Aug 2025 14:18:45 +0530 Donet Tom <donettom@xxxxxxxxxxxxx> wrote:

If register_node() returns an error, it is not handled correctly.
The function will proceed further and try to register CPUs under the
node, which is not correct.

So, in this patch, if register_node() returns an error, we return
immediately from the function.

Fixes: 76b67ed9dce6 ("[PATCH] node hotplug: register cpu: remove node struct")
Signed-off-by: Donet Tom <donettom@xxxxxxxxxxxxx>
---
v1 -> v2
Made the changes based on Oscar’s review comments.

v1 - https://lore.kernel.org/all/20250702112856.295176-1-donettom@xxxxxxxxxxxxx/
---
drivers/base/node.c | 5 +++++
1 file changed, 5 insertions(+)

diff --git a/drivers/base/node.c b/drivers/base/node.c
index c65b4917794e..1608816de67f 100644
--- a/drivers/base/node.c
+++ b/drivers/base/node.c
@@ -883,6 +883,11 @@ int register_one_node(int nid)
node_devices[nid] = node;
error = register_node(node_devices[nid], nid);
+ if (error) {
+ node_devices[nid] = NULL;
+ kfree(node);
+ return error;
+ }
Can this cause a double-free? Looking at register_node(), when
device_register() fails, it calls put_device(&node->dev). The put_device()
call triggers node_device_release() which does kfree(to_node(dev)), freeing
the entire node structure. So when register_node() returns an error, the
node memory is already freed, but this code calls kfree(node) again on the
same memory.

The call chain is: register_node()->device_register() fails->
put_device()->node_device_release()->kfree(to_node(dev)).


Thank you for pointing this out. I will address it and send a v3.



[ This came from automated patch review, but it looks real to me ]

-chris