On Thu, Oct 09, 2014 at 12:01:37PM +0100, Christoffer Dall wrote:

On Wed, Oct 08, 2014 at 10:47:04AM +0100, Catalin Marinas wrote:

quoted

On Tue, Oct 07, 2014 at 08:39:54PM +0100, Christoffer Dall wrote:

quoted

+static inline int kvm_prealloc_hwpgd(struct kvm *kvm, pgd_t *pgd)
+{
+       pud_t *pud;
+       pmd_t *pmd;
+       unsigned int order, i;
+       unsigned long hwpgd;
+
+       if (KVM_PREALLOC_LEVEL == 0)
+               return 0;
+
+       order = get_order(PTRS_PER_S2_PGD);

Isn't order always 0 here? Based on our IRC discussion, PTRS_PER_S2_PGD
is 16 or less and the order should not be used.

no, if the kernel has 4K pages and 4 levels, then PGDIR_SHIFT is 39, and
KVM_PHYS_SHIFT stays 40, so that means PTRS_PER_S2_PGD becomes 2, which
means we concatenate two first level stage-2 page tables, which means we
need to allocate two consecutive pages, giving us an order of 1, not 0.

So if PTRS_PER_S2_PGD is 2, how come get_order(PTRS_PER_S2_PGD) == 1? My
reading of the get_order() macro is that get_order(2) == 0.

Did you mean get_order(PTRS_PER_S2_PGD * PAGE_SIZE)?

Or you could define a PTRS_PER_S2_PGD_SHIFT as (KVM_PHYS_SHIFT -
PGDIR_SHIFT) and use this as the order directly.

quoted

quoted

+       hwpgd = __get_free_pages(GFP_KERNEL | __GFP_ZERO, order);

I assume you need __get_free_pages() for alignment.

yes, would you prefer a comment to that fact?

No, that's fine.

-- 
Catalin