Hi,

On Sun, Jan 05, 2025 at 05:24:48PM +0800, Ma Ke wrote:

When device_add(&dev->dev) failed, calling put_device() to explicitly release dev->dev. Otherwise, it could cause double free problem.

How exactly allegedly missing put would cause double free?

As comment of device_add() says, if device_add() succeeds, you should call device_del() when you want to get rid of it. If device_add() has not succeeded, use only put_device() to drop the reference count.

As explained in the kerneldoc for input_register_device(), in case of the failure caller must call input_free_device() which will do the required "put" as well as will handle devm-allocated input devices properly.

Adding call to put_device() as proposed by this patch will indeed introduce double-free.

Found by code review.

Cc: stable@vger.kernel.org Fixes: 0cd587735205 ("Input: preallocate memory to hold event values") Signed-off-by: Ma Ke make24@iscas.ac.cn

---

drivers/input/input.c | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-)

diff --git a/drivers/input/input.c b/drivers/input/input.c index 7f0477e04ad2..a0a36aa90ecc 100644 --- a/drivers/input/input.c +++ b/drivers/input/input.c @@ -2456,8 +2456,10 @@ int input_register_device(struct input_dev *dev) input_dev_poller_finalize(dev->poller); error = device_add(&dev->dev);

• if (error)

• if (error) {

• put_device(&dev->dev);
  

  goto err_devres_free;
• }

path = kobject_get_path(&dev->dev.kobj, GFP_KERNEL); pr_info("%s as %s\n", -- 2.25.1

Thanks.