Arnd, thanks for the comments.
On 07/29/2014 03:00 PM, Arnd Bergmann wrote:
> On Tuesday 29 July 2014 14:42:31 Stanimir Varbanov wrote:
>> taddr = of_translate_address(dev, addrp);
>> - if (taddr == OF_BAD_ADDR)
>> - return -EINVAL;
>> + /*
>> + * if the address is non-translatable to cpu physical address
>> + * fallback to a IORESOURCE_REG resource.
>> + */
>> + if (taddr == OF_BAD_ADDR) {
>> + memset(r, 0, sizeof(*r));
>> + taddr = of_read_number(addrp, 1);
>> + if (taddr == OF_BAD_ADDR)
>> + return -EINVAL;
>> + r->start = taddr;
>> + r->end = taddr + size - 1;
>> + r->flags = IORESOURCE_REG;
>> + r->name = name ? name : dev->full_name;
>> + return 0;
>> + }
>> +
>
> I don't think that everything returning OF_BAD_ADDR makes sense
> to turn into IORESOURCE_REG. It could be an e.g. invalid DT
> representation, a node with #size-cells=<0>, or it could be
> something that gets translated one or more nodes up in the
> tree before it reaches a bus without a ranges property.
>
> Also, you should not rely on #address-cells being hardcoded
> to <1> above.
>
This was just an example. Of course it has many issues and probaly it is
wrong:) The main goal was to understand does IORESOURCE_REG resource
type and parsing the *reg* properties for non-translatable addresses are
feasible. And also does it acceptable by community and OF platform
maintainers.
> How about modifying of_get_address() rather than
> __of_address_to_resource() instead? You could introduce
> a new of_bus entry for each bus you expect to return
> an IORESOURCE_REG, or you could change of_bus_default_get_flags
> to return IORESOURCE_REG if the parent node has no ranges property
> and is not the root node.
IMO the clearer solution is to introduce a new of_bus bus. In that case
one possible problem will be how to distinguish the non-translatable and
the other buses. Also the *device_type* property is deprecated for non
PCI devices.
The second option will need to change the prototype of .get_flags method
to receive device_node structure.
Thoughts?
--
regards,
Stan
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