Re: [PATCH v8 07/16] x86/virt/tdx: Use all system memory when initializing TDX module as TDX memory

[Dave Hansen]

On 1/10/23 04:09, Huang, Kai wrote:
> On Mon, 2023-01-09 at 08:51 -0800, Dave Hansen wrote:
>> On 1/9/23 03:48, Huang, Kai wrote:
>>>>>>> This can also be enhanced in the future, i.e. by allowing adding non-TDX
>>>>>>> memory to a separate NUMA node.  In this case, the "TDX-capable" nodes
>>>>>>> and the "non-TDX-capable" nodes can co-exist, but the kernel/userspace
>>>>>>> needs to guarantee memory pages for TDX guests are always allocated from
>>>>>>> the "TDX-capable" nodes.
>>>>>
>>>>> Why does it need to be enhanced?  What's the problem?
>>>
>>> The problem is after TDX module initialization, no more memory can be hot-added
>>> to the page allocator.
>>>
>>> Kirill suggested this may not be ideal. With the existing NUMA ABIs we can
>>> actually have both TDX-capable and non-TDX-capable NUMA nodes online. We can
>>> bind TDX workloads to TDX-capable nodes while other non-TDX workloads can
>>> utilize all memory.
>>>
>>> But probably it is not necessarily to call out in the changelog?
>>
>> Let's say that we add this TDX-compatible-node ABI in the future.  What
>> will old code do that doesn't know about this ABI?
> 
> Right.  The old app will break w/o knowing the new ABI.  One resolution, I
> think, is we don't introduce new userspace ABI, but hide "TDX-capable" and "non-
> TDX-capable" nodes in the kernel, and let kernel to enforce always allocating
> TDX guest memory from those "TDX-capable" nodes.

That doesn't actually hide all of the behavior from users.  Let's say
they do:

	numactl --membind=6 qemu-kvm ...

In other words, take all of this guest's memory and put it on node 6.
There lots of free memory on node 6 which is TDX-*IN*compatible.  Then,
they make it a TDX guest:

	numactl --membind=6 qemu-kvm -tdx ...

What happens?  Does the kernel silently ignore the --membind=6?  Or does
it return -ENOMEM somewhere and confuse the user who has *LOTS* of free
memory on node 6.

In other words, I don't think the kernel can just enforce this
internally and hide it from userspace.

>> Is there something fundamental that keeps a memory area that spans two
>> nodes from being removed and then a new area added that is comprised of
>> a single node?
>> Boot time:
>>
>> 	| memblock  |  memblock |
>> 	<--Node=0--> <--Node=1-->
>>
>> Funky hotplug... nothing to see here, then:
>>
>> 	<--------Node=2-------->
> 
> I must have missed something, but how can this happen?
> 
> I had memory that this cannot happen because the BIOS always allocates address
> ranges for all NUMA nodes during machine boot.  Those address ranges don't
> necessarily need to have DIMM fully populated but they don't change during
> machine's runtime.

Is your memory correct?  Is there evidence, or requirements in any
specification to support your memory?