On Tue, Aug 06, 2013 at 05:48:54PM +0200, Jan Kiszka wrote:
> On 2013-08-06 17:04, Zhang, Yang Z wrote:
> > Gleb Natapov wrote on 2013-08-06:
> >> On Tue, Aug 06, 2013 at 02:12:51PM +0000, Zhang, Yang Z wrote:
> >>> Gleb Natapov wrote on 2013-08-06:
> >>>> On Tue, Aug 06, 2013 at 11:44:41AM +0000, Zhang, Yang Z wrote:
> >>>>> Gleb Natapov wrote on 2013-08-06:
> >>>>>> On Tue, Aug 06, 2013 at 10:39:59AM +0200, Jan Kiszka wrote:
> >>>>>>> From: Jan Kiszka <jan.kiszka@xxxxxxxxxxx>
> >>>>>>>
> >>>>>>> If nested EPT is enabled, the L2 guest may change CR3 without any
> >>>>>>> exits. We therefore have to read the current value from the VMCS
> >>>>>>> when switching to L1. However, if paging wasn't enabled, L0 tracks
> >>>>>>> L2's CR3, and GUEST_CR3 rather contains the real-mode identity map.
> >>>>>>> So we need to retrieve CR3 from the architectural state after
> >>>>>>> conditionally updating it - and this is what kvm_read_cr3 does.
> >>>>>>>
> >>>>>> I have a headache from trying to think about it already, but
> >>>>>> shouldn't
> >>>>>> L1 be the one who setups identity map for L2? I traced what
> >>>>>> vmcs_read64(GUEST_CR3)/kvm_read_cr3(vcpu) return here and do not
> >>>>>> see
> >>>>> Here is my understanding:
> >>>>> In vmx_set_cr3(), if enabled ept, it will check whether target
> >>>>> vcpu is enabling
> >>>> paging. When L2 running in real mode, then target vcpu is not
> >>>> enabling paging and it will use L0's identity map for L2. If you
> >>>> read GUEST_CR3 from VMCS, then you may get the L2's identity map
> >>>> not
> >> L1's.
> >>>>>
> >>>> Yes, but why it makes sense to use L0 identity map for L2? I didn't
> >>>> see different vmcs_read64(GUEST_CR3)/kvm_read_cr3(vcpu) values because
> >>>> L0 and L1 use the same identity map address. When I changed identity
> >>>> address L1 configures vmcs_read64(GUEST_CR3)/kvm_read_cr3(vcpu) are
> >>>> indeed different, but the real CR3 L2 uses points to L0 identity map.
> >>>> If I zero L1 identity map page L2 still works.
> >>>>
> >>> If L2 in real mode, then L2PA == L1PA. So L0's identity map also works
> >>> if L2 is in real mode.
> >>>
> >> That not the point. It may work accidentally for kvm on kvm, but what
> >> if other hypervisor plays different tricks and builds different ident map for its guest?
> > Yes, if other hypervisor doesn't build the 1:1 mapping for its guest, it will fail to work. But I cannot imagine what kind of hypervisor will do this and what the purpose is.
> > Anyway, current logic is definitely wrong. It should use L1's identity map instead L0's.
>
> So something like this is rather needed?
>
> diff --git a/arch/x86/kvm/vmx.c b/arch/x86/kvm/vmx.c
> index 44494ed..60a3644 100644
> --- a/arch/x86/kvm/vmx.c
> +++ b/arch/x86/kvm/vmx.c
> @@ -3375,8 +3375,10 @@ static void vmx_set_cr3(struct kvm_vcpu *vcpu, unsigned long cr3)
> if (enable_ept) {
> eptp = construct_eptp(cr3);
> vmcs_write64(EPT_POINTER, eptp);
> - guest_cr3 = is_paging(vcpu) ? kvm_read_cr3(vcpu) :
> - vcpu->kvm->arch.ept_identity_map_addr;
> + if (is_paging(vcpu) || is_guest_mode(vcpu))
> + guest_cr3 = kvm_read_cr3(vcpu) :
> + else
> + guest_cr3 = vcpu->kvm->arch.ept_identity_map_addr;
> ept_load_pdptrs(vcpu);
> }
>
That what I am thinking, will think about it some more tomorrow.
--
Gleb.
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