On Thu, Nov 14, 2019 at 04:08:25PM +1100, Paul Mackerras wrote:
On Wed, Nov 13, 2019 at 01:50:42PM -0800, Ram Pai wrote:
quoted
On Thu, Nov 14, 2019 at 08:18:24AM +1100, Paul Mackerras wrote:
quoted
On Tue, Nov 12, 2019 at 10:32:33PM -0800, Ram Pai wrote:
quoted
On Wed, Nov 13, 2019 at 11:14:27AM +1100, Paul Mackerras wrote:
quoted
On Tue, Nov 12, 2019 at 06:45:55AM -0800, Ram Pai wrote:
quoted
On Tue, Nov 12, 2019 at 10:32:04PM +1100, Paul Mackerras wrote:
quoted
On Mon, Nov 11, 2019 at 11:52:15PM -0800, Ram Pai wrote:
quoted
There is subtle problem removing that code from the assembly.
If the H_SVM_INIT_ABORT hcall returns to the ultravisor without clearing
kvm->arch.secure_guest, the hypervisor will continue to think that the
VM is a secure VM. However the primary reason the H_SVM_INIT_ABORT
hcall was invoked, was to inform the Hypervisor that it should no longer
consider the VM as a Secure VM. So there is a inconsistency there.
Most of the checks that look at whether a VM is a secure VM use code
like "if (kvm->arch.secure_guest & KVMPPC_SECURE_INIT_DONE)". Now
since KVMPPC_SECURE_INIT_ABORT is 4, an if statement such as that will
take the false branch once we have set kvm->arch.secure_guest to
KVMPPC_SECURE_INIT_ABORT in kvmppc_h_svm_init_abort. So in fact in
most places we will treat the VM as a normal VM from then on. If
there are any places where we still need to treat the VM as a secure
VM while we are processing the abort we can easily do that too.
Is the suggestion -- KVMPPC_SECURE_INIT_ABORT should never return back
to the Ultravisor? Because removing that assembly code will NOT lead the
No. The suggestion is that vcpu->arch.secure_guest stays set to
KVMPPC_SECURE_INIT_ABORT until userspace calls KVM_PPC_SVM_OFF.
In the fast_guest_return path, if it finds
(kvm->arch.secure_guest & KVMPPC_SECURE_INIT_ABORT) is true, should it return to
UV or not?
Ideally it should return back to the ultravisor the first time
KVMPPC_SECURE_INIT_ABORT is set, and not than onwards.
What is ideal about that behavior? Why would that be a particularly
good thing to do?
It is following the rule -- "return back to the caller".
That doesn't address the question of why vcpu->arch.secure_guest
should be cleared at the point where we do that.
Ok. here is the sequence that I expect to happen.
-------------------------------------------------------------------
1) VM makes a ESM(Enter Secure Mode) ucall.
1A) UV does the H_SVM_INIT_START hcall... the chit-chat between UV and HV happens
and somewhere in that chit-chat, UV decides that the ESM call
cannot be successfully completed. Hence it calls
H_SVM_INIT_ABORT to inform the Hypervisor.
1A_a) Hypervisor aborts the VM's transition and returns back to the ultravisor.
1B) UV cleans up the state of the VM on its side and returns
back to the VM, with failure. VM is still in a normal VM state.
Its MSR(S) is still 0.
2) VM gets a HDEC exception.
2A) Hypervisor receives that HDEC exception. It handles the exception
and returns back to the VM.
-------------------------------------------------------------------
If you agree with the above sequence than, the patch needs all the proposed changes.
At step (1A_a) and step (2A), the hypervisor is faced with the question
-- Where should the control be returned to?
for step 1A_a it has to return to the UV, and for step 2A it has to
return to the VM. In other words the control has to **return back to
the caller**.
It certainly has to rely on the vcpu->arch.secure_guest flag to do so.
If any flag in vcpu->arch.secure_guest is set, it has to return to
UV. Otherwise it has to return to VM.
This is the reason, the proposed patch in the fast_guest_return path
looks at the vcpu->arch.secure_guest. If it finds any flag set, it
returns to UV. However before returning to UV, it also clears all the
flags if H_SVM_INIT_ABORT is set. This is to ensure that HV does not
return to the wrong place; i.e to UV, but returns to the VM at step 2A.
RP