Re: [Xen-devel] [v3][PATCH 07/16] hvmloader/pci: skip reserved ranges

["Jan Beulich" <JBeulich@xxxxxxxx>]

>>> On 17.06.15 at 09:54, <tiejun.chen@xxxxxxxxx> wrote:
> On 2015/6/17 15:19, Jan Beulich wrote:
>>>>> On 17.06.15 at 09:10, <tiejun.chen@xxxxxxxxx> wrote:
>>> Yeah, this may waste some spaces in this worst case but I this think
>>> this can guarantee our change don't impact on the original expectation,
>>> right?
>>
>> "Some space" may be multiple Gb (e.g. the frame buffer of a graphics
> 
> Sure.
> 
>> card), which is totally unacceptable.
>>
> 
> But then I don't understand what's your way. How can we fit all pci 
> devices just with "the smallest power-of-2 region enclosing the reserved 
> device memory"?
> 
> For example, the whole pci memory is sitting at
> [0xa0000000, 0xa2000000]. And there are two PCI devices, A and B. Note 
> each device needs to be allocated with 0x1000000. So if without 
> concerning RMRR,
> 
> A. [0xa0000000,0xa1000000]
> B. [0xa1000000,0xa2000000]
> 
> But if one RMRR resides at [0xa0f00000, 0xa1f00000] which obviously 
> generate its own alignment with 0x1000000. So the pci memory is expended 
> as [0xa0000000, 0xa3000000], right?
> 
> Then actually the whole pci memory can be separated three segments like,
> 
> #1. [0xa0000000, 0xa0f00000]
> #2. [0xa0f00000, 0xa1f00000] -> RMRR would occupy
> #3. [0xa1f00000, 0xa3000000]
> 
> So just #3 can suffice to allocate but just for one device, right?

Right, i.e. this isn't even sufficient - you need [a0000000,a3ffffff]
to fit everything (but of course you can put smaller BARs into the
unused ranges [a0000000,a0efffff] and [a1f00000,a1ffffff]).
That's why I said it's not going to be tricky to get all corner cases
right _and_ not use up more space than needed.

>>>> ought to work out the smallest power-of-2 region enclosing the
>>>
>>> Okay. I remember the smallest size of a given PCI I/O space is 8 bytes,
>>> and the smallest size of a PCI memory space is 16 bytes. So
>>>
>>> /* At least 16 bytes to align a PCI BAR size. */
>>> uint64_t align = 16;
>>>
>>> reserved_start = memory_map.map[j].addr;
>>> reserved_size = memory_map.map[j].size;
>>>
>>> reserved_start = (reserved_star + align) & ~(align - 1);
>>> reserved_size = (reserved_size + align) & ~(align - 1);
>>>
>>> Is this correct?
>>
>> Simply aligning the region doesn't help afaict. You need to fit it
>> with the other MMIO allocations.
> 
> I guess you're saying just those mmio allocations conflicting with RMRR? 
> But we don't know these exact addresses until we finalize to allocate 
> them, right?

That's the point - you need to allocate them _around_ the reserved
regions.

Jan


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