On 11/03/2015 05:51 PM, Paolo Bonzini wrote:
On 03/11/2015 15:36, Andrey Smetanin wrote:
if I run a patched QEMU but I *do not* enable the synthetic interrupt
controller. I can fix it by wrapping the calls to synic_exit with "if
(!host)", but I haven't checked yet the source---so that may not be the
proper fix. Sorry for not having looked more in detail.
Could you please specify test case(kvm unit tests ?) and kernel/qemu(if
it's not standard)?
It happens just by starting QEMU.
Kernel: kvm/queue
+ kvm/irqchip: kvm_arch_irq_routing_update renaming split
+ kvm/x86: split ioapic-handled and EOI exit bitmaps
+ kvm/x86: Hyper-V synthetic interrupt controller
+ kvm/x86: Hyper-V kvm exit
QEMU: 3a958f559ecd
+ standard-headers/x86: add Hyper-V SynIC constants
+ target-i386/kvm: Hyper-V SynIC MSR's support
+ linux-headers/kvm: add Hyper-V SynIC irq routing type and struct
+ kvm: Hyper-V SynIC irq routing support
+ linux-headers/kvm: KVM_EXIT_HYPERV type and struct
+ target-i386/hyperv: Hyper-V SynIC SINT routing and vCPU exit
+ hw/misc: Hyper-V test device 'hyperv-testdev'
Can be reproduced just with
"../qemu/+build/x86_64-softmmu/qemu-system-x86_64 --enable-kvm -cpu
kvm64 -display none".
Thanks!
We probably found root case -
qemu reads/writes Hyper-V SynIC msrs just by check SynIC MSR's support
in kernel. So KVM synic exits into userspace(at SynIC MSR's writes),
while userspace Hyper-V SynIC handler doesn't expect this exit(cpu
'hv-synic' option is not set), so handler returns -1 and qemu exits.
Paolo
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