Re: Runtime Memory Validation in Intel-TDX and AMD-SNP

[Mike Rapoport <rppt@kernel.org>]

On Sun, Jul 25, 2021 at 09:28:28PM +0300, Kirill A. Shutemov wrote:
> On Sun, Jul 25, 2021 at 12:16:45PM +0300, Mike Rapoport wrote:
> > On Fri, Jul 23, 2021 at 07:29:59PM +0300, Kirill A. Shutemov wrote:
> > > On Fri, Jul 23, 2021 at 06:23:39PM +0300, Mike Rapoport wrote:
> > > > > @@ -1318,9 +1327,14 @@ void __init e820__memblock_setup(void)
> > > > >  		if (entry->type == E820_TYPE_SOFT_RESERVED)
> > > > >  			memblock_reserve(entry->addr, entry->size);
> > > > >  
> > > > > -		if (entry->type != E820_TYPE_RAM && entry->type != E820_TYPE_RESERVED_KERN)
> > > > > +		if (entry->type != E820_TYPE_RAM &&
> > > > > +		    entry->type != E820_TYPE_RESERVED_KERN &&
> > > > > +		    entry->type != E820_TYPE_UNACCEPTED)
> > > > >  			continue;
> > > > 
> > > > If I understand correctly, you assume that
> > > > 
> > > > * E820_TYPE_RAM and E820_TYPE_RESERVED_KERN regions are already accepted by
> > > >   firmware/booloader
> > > > * E820_TYPE_UNACCEPTED would have been E820_SYSTEM_RAM if we'd disabled
> > > >   encryption
> > > > 
> > > > What happens with other types? Particularly E820_TYPE_ACPI and
> > > > E820_TYPE_NVS that may reside in memory and might have been accepted by
> > > > BIOS.
> > > 
> > > Any accessible memory that not marked as UNACCEPTED has to be accepted
> > > before kernel gets control.
> > 
> > Hmm, that would mean that everything that runs before the kernel must
> > maintain precise E820 map. If we use 2M chunk as basic unit for accepting
> > memory, the firmware must also use the same basic unit. E.g. we can't have
> > an ACPI table squeezed between E820_TYPE_UNACCEPTED.
> 
> No. See mark_unaccepted(). Any chunks that cannot be accepted with 2M, get
> accepted upfront, so we will not need to track them.

What will happen with the following E820 table:

0x400000 - 0x401000 - ACPI (accepted by BIOS)
0x401000 - 0x408000 - UNACCEPTED
0x408000 - 0x409000 - ACPI (accepted by BIOS)

> (I've just realized that mark_unaccepted() is buggy if 'start' and 'end'
> are in the same 2M. Will fix.)
> 
 
> > > > > --- a/mm/memblock.c
> > > > > +++ b/mm/memblock.c
> > > > > @@ -814,6 +814,7 @@ int __init_memblock memblock_reserve(phys_addr_t base, phys_addr_t size)
> > > > >  	memblock_dbg("%s: [%pa-%pa] %pS\n", __func__,
> > > > >  		     &base, &end, (void *)_RET_IP_);
> > > > >  
> > > > > +	accept_pages(base, base + size);
> > > > 
> > > > Hmm, I'm not sure memblock_reserve() is the right place to accept pages. It
> > > > can be called to reserve memory owned by firmware which not necessarily
> > > > would be encrypted. Besides, memblock_reserve() may be called for absent
> > > > memory, could be it'll confuse TDX/SEV?
> > > 
> > > Such memory will not be marked as unaccepted and accept_pages() will do
> > > nothing.
> > > 
> > > > Ideally, the call to accept_pages() should live in
> > > > memblock_alloc_range_nid(), but unfortunately there still stale
> > > > memblock_find_in_range() + memblock_reserve() pairs in x86 setup code.
> > > 
> > > memblock_reserve() is the root of memory allocation in the early boot and
> > > it is natual place to do the trick. Unless we have a good reason to move
> > > it somewhere I would keep it here.
>
> > I think it is better to accept memory that is actually allocated rather
> > than marked as being used. It'll make it more robust against future changes
> > in memblock_reserve() callers and in what is accept_pages() in your patch. 
> 
> I disagree.
> 
> If we move accept_pages() up to callers we will make less robust: any new
> user of memblock_reserve() has to consider if accept_pages() is needed and
> like would ignore it since it's not essential for any non-TDX/non-SEV use
> case.

I do not suggest to move accept_pages() to all the callers of
memblock_reserve(). I suggest to replace memblock_find_in_range() +
memblock_reserve() pairs with an appropriate memblock_alloc call, make
memblock_find_in_range() static and put accept_pages() there.

This essentially makes memblock_find_in_range() the root of early memory
*allocations* while memblock_reserve() would be only used to mark the
memory that is already used before the allocations can start.

Then we only deal with acceptance of the memory kernel actually allocates.

I can't think now of a concrete example of what may go wrong with calling
accept_pages() from memblock_reserve(), it's more of a gut feeling.

-- 
Sincerely yours,
Mike.