On 02/17/19 at 09:25am, Kees Cook wrote:
> On Sat, Feb 16, 2019 at 6:04 AM Baoquan He <b...@redhat.com> wrote:
> >
> > Vmemmap region has different maximum size depending on paging mode.
> > Now its size is hardcoded as 1TB in memory KASLR, this is not
> > right for 5-level paging mode. It will cause overflow if vmemmap
> > region is randomized to be adjacent to cpu_entry_area region and
> > its actual size is bigger than 1TB.
> >
> > So here calculate how many TB by the actual size of vmemmap region
> > and align up to 1TB boundary.
> >
> > Signed-off-by: Baoquan He <b...@redhat.com>
> > ---
> > arch/x86/mm/kaslr.c | 11 ++++++++++-
> > 1 file changed, 10 insertions(+), 1 deletion(-)
> >
> > diff --git a/arch/x86/mm/kaslr.c b/arch/x86/mm/kaslr.c
> > index 97768df923e3..ca12ed4e5239 100644
> > --- a/arch/x86/mm/kaslr.c
> > +++ b/arch/x86/mm/kaslr.c
> > @@ -101,7 +101,7 @@ static __initdata struct kaslr_memory_region {
> > } kaslr_regions[] = {
> > { &page_offset_base, 0 },
> > { &vmalloc_base, 0 },
> > - { &vmemmap_base, 1 },
> > + { &vmemmap_base, 0 },
> > };
> >
> > /*
> > @@ -121,6 +121,7 @@ void __init kernel_randomize_memory(void)
> > unsigned long rand, memory_tb;
> > struct rnd_state rand_state;
> > unsigned long remain_entropy;
> > + unsigned long vmemmap_size;
> >
> > vaddr_start = pgtable_l5_enabled() ? __PAGE_OFFSET_BASE_L5 :
> > __PAGE_OFFSET_BASE_L4;
> > vaddr = vaddr_start;
> > @@ -152,6 +153,14 @@ void __init kernel_randomize_memory(void)
> > if (memory_tb < kaslr_regions[0].size_tb)
> > kaslr_regions[0].size_tb = memory_tb;
> >
> > + /*
> > + * Calculate how many TB vmemmap region needs, and align to
> > + * 1TB boundary.
>
> Can you describe why this is the right calculation? (This will help
> explain why 4-level is different from 5-level here.)
In the old code, the size of vmemmap is hardcoded as 1 TB. This is true
in 4-level paging mode, 64 TB RAM supported at most, and usually
sizeof(struct page) is 64 Bytes, it happens to be 1 TB.
However, in 5-level paging mode, 4 PB is the biggest RAM size we can
support, it's (4 PB)/64 == 1<<48, namely 256 TB area needed for vmemmap,
assuming sizeof(struct page) is 64 Bytes here.
So, the hardcoding of 1 TB is not correct for 5-level paging mode.
Thanks
Baoquan
>
> > + */
> > + vmemmap_size = (kaslr_regions[0].size_tb << (TB_SHIFT -
> > PAGE_SHIFT)) *
> > + sizeof(struct page);
> > + kaslr_regions[2].size_tb = DIV_ROUND_UP(vmemmap_size, 1UL <<
> > TB_SHIFT);
> > +
