On Thu, Jul 15, 2021 at 07:10:54PM +0100, Matthew Wilcox wrote:
> On Thu, Jul 15, 2021 at 05:47:41PM +0100, Russell King (Oracle) wrote:
> > On Thu, Jul 15, 2021 at 02:46:10PM +0100, Matthew Wilcox (Oracle) wrote:
> > > This is the order of the page table allocation, not the order of a PMD.
> > > -#define PMD_ORDER 3
> > > +#define PMD_TABLE_ORDER 3
> > > #else
> > > #define PG_DIR_SIZE 0x4000
> > > -#define PMD_ORDER 2
> > > +#define PMD_TABLE_ORDER 2
> >
> > I think PMD_ENTRY_ORDER would make more sense here - this is the
> > power-of-2 of an individual PMD entry, not of the entire table.
>
> But ... we have two kinds of PMD entries. We have the direct entry that
> points to a 1-16MB sized chunk of memory, and we have the table entry that
> points to a 4k-32k chunk of memory that contains PTEs. So I don't think
> calling it 'entry' order actually disambiguates anything. That's why
> I went with 'table' -- I can't think of anything else to call it!
> PMD_PTE_ARRAY_ORDER doesn't seem like an improvement to me ...
There may be two kinds of PMD entries, but that isn't relevant here.
Going back to the original terminology, 1 << PMD_ORDER here is the
size of each PMD entry. It doesn't have anything to do with how much
memory is being mapped by each entry.
I think what is confusing you is stuff like:
add r0, r4, #KERNEL_OFFSET >> (SECTION_SHIFT - PMD_ORDER)
r4 is the base address of the page tables, and r0 is the address of
the entry we want to manipulate for "KERNEL_OFFSET" - which is the
virtual address. 1 << SECTION_SHIFT is how much memory each entry
maps (and this is fixed here - there's no variability as you suggest
above.)
Effectively, the calculation above is:
index = KERNEL_OFFSET >> SECTION_SHIFT;
pmd_entry_size = 1 << PMD_ORDER;
r0 = base + index * pmd_entry_size;
but in a single instruction as we can be sure that KERNEL_OFFSET will
have zeros as the low bits after shifting by SECTION_SHIFT - PMD_ORDER.
Hope this helps to explain what this PMD_ORDER is actually doing here.
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