Re: [PATCH 21/38] workqueue: add missing put_device call

Levente Kurusa
2013-12-19T15:31:42Z

On 12/19/2013 04:20 PM, Tejun Heo wrote:
> Umm... this doesn't look right. You're basically converting the code
> to the following,
>
> x = kmalloc();
> if (register(x) < 0)
> put(x)
>
> They're not symmetrical anymore. register(), or any API call really,
> isn't supposed to have side effects which need explicit cleanup after
> a failure. The fact that x is properly initialized even after
> register(x) failed is a coincidental implementation detail which
> shouldn't be depended upon. Your patch is actively breaking the
> convention for no good reason.
>
> Nacked-by: Tejun Heo <tj@kernel.org>
>
>>From the patch title, I suppose you posted a bunch of patches towards
> this direction. Please consider all of them nacked if they're doing
> the same thing.
>
> Thanks.
>

The reason I removed the kfree() was because the put_device() will decrement
wq_dev->dev's reference count to zero (it is set to one by device_register) and hence the
wq_device_release() will be called. Now, this effectively does the same the kfree() call
would have done but also driver core is notified.

Also, if you take a look at the comment for the device_register() function, it explicitly
says NOT to kfree the struct device, but instead call put_device() and let the device's release()
function take care.

--
Regards,
Levente Kurusa