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Re: [RFC PATCH v2 32/44] tdx: add kvm_tdx_enabled() accessor for later u

From: Connor Kuehl
Subject: Re: [RFC PATCH v2 32/44] tdx: add kvm_tdx_enabled() accessor for later use
Date: Thu, 22 Jul 2021 12:53:34 -0500
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:78.0) Gecko/20100101 Thunderbird/78.11.0

On 7/7/21 7:55 PM, isaku.yamahata@gmail.com wrote:
From: Isaku Yamahata <isaku.yamahata@intel.com>

Signed-off-by: Isaku Yamahata <isaku.yamahata@intel.com>
  include/sysemu/tdx.h  | 1 +
  target/i386/kvm/kvm.c | 5 +++++
  2 files changed, 6 insertions(+)

diff --git a/include/sysemu/tdx.h b/include/sysemu/tdx.h
index 70eb01348f..f3eced10f9 100644
--- a/include/sysemu/tdx.h
+++ b/include/sysemu/tdx.h
@@ -6,6 +6,7 @@
  #include "hw/i386/pc.h"
bool kvm_has_tdx(KVMState *s);
+bool kvm_tdx_enabled(void);
  int tdx_system_firmware_init(PCMachineState *pcms, MemoryRegion *rom_memory);
diff --git a/target/i386/kvm/kvm.c b/target/i386/kvm/kvm.c
index af6b5f350e..76c3ea9fac 100644
--- a/target/i386/kvm/kvm.c
+++ b/target/i386/kvm/kvm.c
@@ -152,6 +152,11 @@ int kvm_set_vm_type(MachineState *ms, int kvm_type)
      return -ENOTSUP;
+bool kvm_tdx_enabled(void)
+    return vm_type == KVM_X86_TDX_VM;

Is this the whole story? Does this guarantee that the VM QEMU is
responsible to bring up is a successfully initialized TD?

From my reading of the series as it unfolded, this looks like the
function proves that KVM can support TDs and that the user requested
a TDX kvm-type, not that we have a fully-formed TD.

Is it possible to associate this with a more verifiable metric that
the TD has been or will be created successfully? I.e., once the VM
has successfully called the TDX INIT ioctl or has finalized setup?

My question mainly comes from a later patch in the series, where the
"query-tdx-capabilities" and "query-tdx" QMP commands are added.

Forgive me if I am misinterpreting the semantics of each of these

"query-tdx-capabilities" sounds like it answers the question of
"can it run a TD?"

and "query-tdx" sounds like it answers the question of "is it a TD?"

Is the assumption with "query-tdx" that anything that's gone wrong
with developing a TD will have resulted in the QEMU process exiting
and therefore if we get to a point where we can run "query-tdx" then
we know the TD was successfully formed?

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