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

[David Hildenbrand]

On 18.09.25 07:41, Donet Tom wrote:
> 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@xxxxxxxxxxxxx>
> ---
>   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;
>   	}
>   

Yes, that matches what other users (staring at mm/memory-tiers.c) do.

I wonder if we should just inline register_node() into register_one_node().

Then it's clearer that we perform a put_device() already in there.

On top of that, we could then just s/register_one_node/register_node/

And then we could do a similar cleanup for unregister_one_node / 
unregister_node where I don't consider the split function really valuable.

--
Cheers

David / dhildenb