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

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Message-ID: <20190218095010.GJ14858@MiWiFi-R3L-srv>
Date:   Mon, 18 Feb 2019 17:50:10 +0800
From:   Baoquan He <bhe@...hat.com>
To:     Kees Cook <keescook@...omium.org>
Cc:     LKML <linux-kernel@...r.kernel.org>,
        Thomas Gleixner <tglx@...utronix.de>,
        Ingo Molnar <mingo@...hat.com>, Borislav Petkov <bp@...en8.de>,
        "H. Peter Anvin" <hpa@...or.com>,
        Dave Hansen <dave.hansen@...ux.intel.com>,
        Andy Lutomirski <luto@...nel.org>,
        Peter Zijlstra <peterz@...radead.org>, X86 ML <x86@...nel.org>,
        Mike Travis <travis@....com>,
        Thomas Garnier <thgarnie@...gle.com>,
        Andrew Morton <akpm@...ux-foundation.org>,
        Masahiro Yamada <yamada.masahiro@...ionext.com>,
        "Kirill A. Shutemov" <kirill@...temov.name>
Subject: Re: [PATCH v3 5/6] x86/mm/KASLR: Calculate the actual size of
 vmemmap region

On 02/17/19 at 09:25am, Kees Cook wrote:
> On Sat, Feb 16, 2019 at 6:04 AM Baoquan He <bhe@...hat.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@...hat.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);
> > +