On Sep 2, 2015 5:46 AM, "Jan Beulich" <jbeul...@suse.com> wrote:
>
> While commit 37868fe113 ("x86/ldt: Make modify_ldt synchronous") added
> a nice comment explaining that Xen needs page-aligned whole page chunks
> for guest descriptor tables, it then nevertheless used kzalloc() on the
> small size path. As I'm unaware of guarantees for kmalloc(PAGE_SIZE, )
> to return page-aligned memory blocks, I believe this needs to be
> switched back to __get_free_page().
>
> Signed-off-by: Jan Beulich <jbeul...@suse.com>
> Cc: Andy Lutomirski <l...@kernel.org>
> Cc: Boris Ostrovsky <boris.ostrov...@oracle.com>
> Cc: David Vrabel <david.vra...@citrix.com>
> Cc: Konrad Rzeszutek Wilk <konrad.w...@oracle.com>
> ---
> v2: Also adjust the freeing side.
> ---
> arch/x86/kernel/ldt.c | 4 ++--
> 1 file changed, 2 insertions(+), 2 deletions(-)
>
> --- 4.2/arch/x86/kernel/ldt.c
> +++ 4.2-x86-LDT-alloc/arch/x86/kernel/ldt.c
> @@ -58,7 +58,7 @@ static struct ldt_struct *alloc_ldt_stru
> if (alloc_size > PAGE_SIZE)
> new_ldt->entries = vzalloc(alloc_size);
> else
> - new_ldt->entries = kzalloc(PAGE_SIZE, GFP_KERNEL);
> + new_ldt->entries = (void
> *)__get_free_page(GFP_KERNEL|__GFP_ZERO);
>
> if (!new_ldt->entries) {
> kfree(new_ldt);
> @@ -95,7 +95,7 @@ static void free_ldt_struct(struct ldt_s
> if (ldt->size * LDT_ENTRY_SIZE > PAGE_SIZE)
> vfree(ldt->entries);
> else
> - kfree(ldt->entries);
> + put_page(virt_to_page(ldt->entries));
FWIW, I'm not convinced this is or was correct. Using free_page looks
a bit safer, and free_page does more than just put_page.
Does anyone here know the right way to do this?
--Andy
> kfree(ldt);
> }
>
>
>
>
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