On Wed, Apr 01, 2009 at 11:40:30AM +0300, Avi Kivity wrote:
> Marcelo Tosatti wrote:
>> On Sun, Mar 29, 2009 at 03:24:08PM +0300, Avi Kivity wrote:
>>
>>>> static int mmu_pages_next(struct kvm_mmu_pages *pvec,
>>>> struct mmu_page_path *parents,
>>>> int i)
>>>> {
>>>> int n;
>>>>
>>>> for (n = i+1; n < pvec->nr; n++) {
>>>> struct kvm_mmu_page *sp = pvec->page[n].sp;
>>>>
>>>> if (sp->role.level == PT_PAGE_TABLE_LEVEL) {
>>>> parents->idx[0] = pvec->page[n].idx;
>>>> return n;
>>>> }
>>>>
>>>> parents->parent[sp->role.level-2] = sp;
>>>> parents->idx[sp->role.level-1] = pvec->page[n].idx;
>>>> }
>>>>
>>>> return n;
>>>> }
>>>>
>>> Do we need to break out of the loop if we switch parents during the
>>> loop (since that will give us a different mmu_page_path)? Or are
>>> callers careful to only pass pvecs which belong to the same shadow
>>> page?
>>>
>>
>> This function builds mmu_page_path for a number of pagetable (leaf)
>> pages. Whenever the path changes, mmu_page_path will be rebuilt.
>>
>> The pages in the pvec must be organized as follows:
>>
>> level4, level3, level2, level1, level1, level1, ...., level3, level2,
>> level1, level1, ...
>>
>> So you don't have to repeat higher levels for a number of leaf pages.
>>
>
> I'm still missing something. That if () tests for level ==
> PT_PAGE_TABLE_LEVEL. So it looks like we'll have batch sizes of 4, 1,
> 1, 1, ... 3, 1, 1, 1, ...?
The input is the pvec array, organized as follows: 4, 3, 2, 1, 1, 1,
2, 1, 1, 3, 2, 1.
The output will be:
1 (pvec position 4), 1 (pos 5), 1 (pos 6). All of them with the same
path.
Then 1 (pos 8), 1 (pos 9). With the same path as before but level 2
being the page in position 7.
So the if() tests for level == PT_PAGE_TABLE_LEVEL because these are the
pages we're interested in walking (the ones that can be unsync). If
we're not PT_PAGE_TABLE_LEVEL, we walk the pvec building mmu_page_path.
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