Re: [PATCH 18/28] PCI/sparc: Use pci_scan_root_bridge() for simplicity

Yijing Wang
2015-01-19T10:54:50Z

On 2015/1/19 16:44, Arnd Bergmann wrote:
> On Monday 19 January 2015 11:17:02 Yijing Wang wrote:
>> On 2015/1/16 18:01, Arnd Bergmann wrote:
>>> On Friday 16 January 2015 09:44:16 Yijing Wang wrote:
>>>> +static void pci_host_bridge_probe_mode(
>>>> + struct pci_host_bridge *host)
>>>> +{
>>>> + host->of_scan = true;
>>>> +}
>>>>
>>>
>>> I probably missed something here, but where does host->of_scan
>>> get used?
>>>
>>
>> It is used in __pci_scan_root_bus() to detect whether of scan bus is needed in patch 11.
>
> Ok, I see it now.
>
>> @@ -2051,10 +2053,17 @@ static struct pci_bus *__pci_scan_root_bus(
>> pci_bus_insert_busn_res(b, host->busnum, 255);
>> }
>>
>> - max = pci_scan_child_bus(b);
>> + if (host->ops && host->ops->phb_probe_mode)
>> + host->ops->phb_probe_mode(host);
>>
>> - if (!found)
>> - pci_bus_update_busn_res_end(b, max);
>> + if (host->of_scan) {
>> + if (host->ops &&host->ops->phb_of_scan_bus)
>> + host->ops->phb_of_scan_bus(host);
>> + } else {
>> + max = pci_scan_child_bus(b);
>> + if (!found)
>> + pci_bus_update_busn_res_end(b, max);
>> + }
>>
>
> I think we can simplify this based on the knowledge that this code
> will remain powerpc specific. How about removing the phb_probe_mode()
> callback and replacing it with just a phb_scan_bus() callback that
> will always get set on sparc, and do this on powerpc:
>
> +static void pci_host_bridge_of_scan_bus(struct pci_host_bridge *host)
> +{
> + int mode = PCI_PROBE_NORMAL;
> + struct pci_bus *bus = host->bus;
> + struct pci_controller *hose = dev_get_drvdata(&host->dev);
> +
> + /* Get probe mode and perform scan */
> + if (hose->dn && ppc_md.pci_probe_mode)
> + mode = ppc_md.pci_probe_mode(bus);
> +
> + pr_debug(" probe mode: %d\n", mode);
> + if (mode == PCI_PROBE_DEVTREE) {
> + of_scan_bus(host->dn, bus);
> + return max; // XXX need to get this from somewhere
> + }
> + return pci_scan_child_bus(b);
> +}
>
> Would that work?

This looks good to me, it is more simple.

Thanks!
Yijing.

> Arnd
>
> .
>

--
Thanks!
Yijing