>>> On 06.06.18 at 16:16, <[email protected]> wrote:
> On 06/06/18 14:50, Jan Beulich wrote:
>>>>> On 04.06.18 at 15:59, <[email protected]> wrote:
>>> --- a/xen/arch/x86/hvm/vmx/vmx.c
>>> +++ b/xen/arch/x86/hvm/vmx/vmx.c
>>> @@ -3696,6 +3696,7 @@ void vmx_vmexit_handler(struct cpu_user_regs *regs)
>>> */
>>> __vmread(EXIT_QUALIFICATION, &exit_qualification);
>>> HVMTRACE_1D(TRAP_DEBUG, exit_qualification);
>>> + __restore_debug_registers(v);
>>> write_debugreg(6, exit_qualification | DR_STATUS_RESERVED_ONE);
>> The change is certainly correct as is, but I'd still like to put out for
>> discussion the alternative option:
>>
>> if ( v->arch.hvm_vcpu.flag_dr_dirty )
>> write_debugreg(6, exit_qualification | DR_STATUS_RESERVED_ONE);
>> else
>> v->arch.debugreg[6] = exit_qualification | DR_STATUS_RESERVED_ONE;
>>
>> After all the guest may know it's single stepping, and may not care to
>> read DR6 at all.
>
> All of this code changes across the series (so this specific suggestion
> is incorrect), but to the recommendation in general...
While I've not made it through the second half of the series yet,
another consideration: To avoid the double DR6 write, yet still avoid
an immediate further exit to restore debug registers, why not
if ( v->arch.hvm_vcpu.flag_dr_dirty )
write_debugreg(6, exit_qualification | DR_STATUS_RESERVED_ONE);
else
{
v->arch.debugreg[6] = exit_qualification | DR_STATUS_RESERVED_ONE;
__restore_debug_registers(v);
}
Jan
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