[RFC PATCH v7 5/8] ice: implement dpll interface to control cgu

[Jiri Pirko]

Thu, May 18, 2023 at 06:06:03PM CEST, arkadiusz.kubalewski at intel.com wrote:
>>From: Jiri Pirko <jiri at resnulli.us>
>>Sent: Tuesday, May 16, 2023 8:26 AM
>>
>>Tue, May 16, 2023 at 12:07:57AM CEST, arkadiusz.kubalewski at intel.com wrote:
>>>>From: Jiri Pirko <jiri at resnulli.us>
>>>>Sent: Wednesday, May 3, 2023 2:19 PM
>>>>
>>>>Fri, Apr 28, 2023 at 02:20:06AM CEST, vadfed at meta.com wrote:
>>>>>From: Arkadiusz Kubalewski <arkadiusz.kubalewski at intel.com>

[...]


>>>>>+			pins[i].pin = NULL;
>>>>>+			return -ENOMEM;
>>>>>+		}
>>>>>+		if (cgu) {
>>>>>+			ret = dpll_pin_register(pf->dplls.eec.dpll,
>>>>>+						pins[i].pin,
>>>>>+						ops, pf, NULL);
>>>>>+			if (ret)
>>>>>+				return ret;
>>>>>+			ret = dpll_pin_register(pf->dplls.pps.dpll,
>>>>>+						pins[i].pin,
>>>>>+						ops, pf, NULL);
>>>>>+			if (ret)
>>>>>+				return ret;
>>>>
>>>>You have to call dpll_pin_unregister(pf->dplls.eec.dpll, pins[i].pin, ..)
>>>>here.
>>>>
>>>
>>>No, in case of error, the caller releases everything
>>ice_dpll_release_all(..).
>>
>>
>>How does ice_dpll_release_all() where you failed? If you need to
>>unregister one or both or none? I know that in ice you have odd ways to
>>handle error paths in general, but this one clearly seems to be broken.
>>
>
>It doesn't have to, as release all would release all anyway.
>Leaving it for now.

So you call dpll_pin_unregister() even for the pin that was not
registered before? How is that even remotely correct?

Fix your error paths, please. I don't understand the resistance here :)

[...]