Re: [PATCH v3 06/16] x86/virt/tdx: Improve PAMT refcounters allocation for sparse memory

Binbin Wu
2025-09-24T08:57:20Z

On 9/24/2025 2:50 PM, Huang, Kai wrote:
> On Tue, 2025-09-23 at 17:38 +0800, Binbin Wu wrote:
>>>> +/*
>>>> + * Allocate PAMT reference counters for the given PFN range.
>>>> + *
>>>> + * It consumes 2MiB for every 1TiB of physical memory.
>>>> + */
>>>> +static int alloc_pamt_refcount(unsigned long start_pfn, unsigned long end_pfn)
>>>> +{
>>>> + unsigned long start, end;
>>>> +
>>>> + start = (unsigned long)tdx_find_pamt_refcount(PFN_PHYS(start_pfn));
>>>> + end   = (unsigned long)tdx_find_pamt_refcount(PFN_PHYS(end_pfn + 1));
>>> (sorry didn't notice this in last version)
>>>
>>> I don't quite follow why we need "end_pfn + 1" instead of just "end_pfn"?
>>>
>>> IIUC this could result in an additional 2M range being populated
>>> unnecessarily when the end_pfn is 2M aligned.
>> IIUC, this will not happen.
>> The +1 page will be converted to 4KB, and will be ignored since in
>> tdx_find_pamt_refcount() the address is divided by 2M.
>>
>> To handle the address unaligned to 2M, +511 should be used instead of +1?
> OK. Thanks for catching. But I still don't get why we need end_pfn + 1.
>
> Also, when end_pfn == 511, would this result in the second refcount being
> returned for the @end, while the intention should be the first refcount?
>
> For example, assuming we have a range [0, 2M), we only need one refcount.
> And the PFN range (which comes from for_each_mem_pfn_range()) would be:
>
> start_pfn == 0
> end_pfn == 512
>
> This will results in @start pointing to the first refcount and @end
> pointing to the second, IIUC.
>
> So it seems we need:
>
> start = (unsigned long)tdx_find_pamt_refcount(PFN_PHYS(start_pfn));
> end = (unsigned long)tdx_find_pamt_refcount(PFN_PHYS(end_pfn) - 1));
> start = round_down(start, PAGE_SIZE);
> end = round_up(end, PAGE_SIZE);

Checked again, this seems to be the right version.

>
> ?