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.kis...@siemens.com> >>>>>> >>>>>> 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.
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