Re: What happens on an INT80 instruction

[Anthony Liguori <anthony-rdkfGonbjUSkNkDKm+mE6A@public.gmane.org>]

Jun Koi wrote:
> On 10/1/07, Anthony Liguori <anthony-rdkfGonbjUSkNkDKm+mE6A@public.gmane.org> wrote:
>   
>> Cameron Macdonell wrote:
>>     
>>> Hi,
>>>
>>> I'm trying to understand guest virtualization at the lower levels.  I
>>> have a somewhat basic question:  How does KVM virtualize an int80
>>> instruction from a guest?  A pointer to an answer is just as good as
>>> an answer itself.
>>>
>>>       
>> The same thing happens as it does on normal hardware.
>>
>> The way VT/SVM works (at a high level), is that certain instructions and
>> events check a special area called the VMCS/VMCB to determine whether
>> the event should generate a vmexit which is really just a special type
>> of trap.
>>
>> There are no hooks for interrupts 32-255 so the hardware operates as it
>> normally would.  If you're interested in getting a trap for int80 within
>> KVM, you'll have to trap sidt/lidt and virtualize the IDT.  You'll need
>> to setup a fake IDT and have the int80 handler do a hypercall.  This is
>> complicated if the guest is using a fast-syscall mechanism.  It may be a
>> little challenging finding a piece of guest memory to take over that has
>> a valid virtual mapping.
>>     
>
> This is a bit vague to me. Why do you need "a piece of guest memory" here?
>   

You don't just need guest memory, you need a valid guest virtual address 
too.  The IDTR contains a guest VA.  If you want to create your own IDT, 
then it has to be a valid VA in the guest's address space.

Regards,

Anthony Liguori

> Thanks,
> Jun
>
>   


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