Re: [PATCH v3 5/6] x86/mm/KASLR: Calculate the actual size of vmemmap region

From: Baoquan He <bhe@redhat.com>
To: Kees Cook <keescook@chromium.org>
Cc: LKML <linux-kernel@vger.kernel.org>,
	Thomas Gleixner <tglx@linutronix.de>,
	Ingo Molnar <mingo@redhat.com>, Borislav Petkov <bp@alien8.de>,
	"H. Peter Anvin" <hpa@zytor.com>,
	Dave Hansen <dave.hansen@linux.intel.com>,
	Andy Lutomirski <luto@kernel.org>,
	Peter Zijlstra <peterz@infradead.org>, X86 ML <x86@kernel.org>,
	Mike Travis <travis@sgi.com>,
	Thomas Garnier <thgarnie@google.com>,
	Andrew Morton <akpm@linux-foundation.org>,
	Masahiro Yamada <yamada.masahiro@socionext.com>,
	"Kirill A. Shutemov" <kirill@shutemov.name>
Subject: Re: [PATCH v3 5/6] x86/mm/KASLR: Calculate the actual size of vmemmap region
Date: Mon, 18 Feb 2019 18:11:33 +0800	[thread overview]
Message-ID: <20190218101133.GL14858@MiWiFi-R3L-srv> (raw)
In-Reply-To: <20190218095010.GJ14858@MiWiFi-R3L-srv>

On 02/18/19 at 05:50pm, Baoquan He wrote:
> On 02/17/19 at 09:25am, Kees Cook wrote:
> > On Sat, Feb 16, 2019 at 6:04 AM Baoquan He <bhe@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 <bhe@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,

Sorry, this should be (4 PB)/64 == 1<<46, 64 TB is the maximum size 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);
> > > +