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

[Huang, Kai]

On Thu, 2023-01-12 at 08:56 +0800, Huang, Ying wrote:
> "Huang, Kai" <kai.huang@xxxxxxxxx> writes:
> 
> > On Tue, 2023-01-10 at 08:18 -0800, Hansen, Dave wrote:
> > > 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.
> > 
> > IIUC, the kernel, for instance KVM who has knowledge the 'task_struct' is a TDX
> > guest, can manually AND "TDX-capable" node masks to task's mempolicy, so that
> > the memory will always be allocated from those "TDX-capable" nodes.  KVM can
> > refuse to create the TDX guest if it found task's mempolicy doesn't have any
> > "TDX-capable" node, and print out a clear message to the userspace.
> > 
> > But I am new to the core-mm, so I might have some misunderstanding.
> 
> KVM here means in-kernel KVM module?  If so, KVM can only output some
> message in dmesg.  Which isn't very good for users to digest.  It's
> better for the user space QEMU to detect whether current configuration
> is usable and respond to users, via GUI, or syslog, etc.

I am not against this. For instance, maybe we can add some dedicated error code
and let KVM return it to Qemu, but I don't want to speak for KVM guys.  We can
discuss this more when we have patches actually sent out to the community.