Re: [PATCH v4 0/6] Nonatomic interrupt injection

[Marcelo Tosatti <mtosatti@redhat.com>]

On Thu, Sep 16, 2010 at 03:35:19PM +0200, Avi Kivity wrote:
>  On 08/30/2010 02:36 PM, Avi Kivity wrote:
> >This patchset changes interrupt injection to be done from normal process
> >context instead of interrupts disabled context.  This is useful for real
> >mode interrupt injection on Intel without the current hacks (injecting as
> >a software interrupt of a vm86 task), reducing latencies, and later, for
> >allowing nested virtualization code to use kvm_read_guest()/kvm_write_guest()
> >instead of kmap() to access the guest vmcb/vmcs.
> >
> >Seems to survive a hack that cancels every 16th entry, after injection has
> >already taken place.
> >
> >With the PIC reset fix posted earlier, this passes autotest on both AMD and
> >Intel, with in-kernel irqchip.  I'll run -no-kvm-irqchip tests shortly.
> >
> >Please review carefully, esp. the first patch.  Any missing kvm_make_request()
> >there may result in a hung guest.
> >
> 
> This is now merged, with the change pointed out by Marcelo.  Windows
> XP x64 fails installation without
> 
> (vmx.c handle_cr())
>          case 8: {
>                  u8 cr8_prev = kvm_get_cr8(vcpu);
>                  u8 cr8 = kvm_register_read(vcpu, reg);
>                  kvm_set_cr8(vcpu, cr8);
>                  skip_emulated_instruction(vcpu);
>                  if (irqchip_in_kernel(vcpu->kvm))
>                      return 1;
> -                if (cr8_prev <= cr8)
> -                    return 1;
>                  vcpu->run->exit_reason = KVM_EXIT_SET_TPR;
>                  return 0;
>              }
> 
> Which doesn't make any sense (anyone?).  The failure is present even
> without the patchset, and is fixed by the same hack, so a regression
> was not introduced.

If userspace does not have an uptodate TPR value, it can signal an
interrupt that is now blocked? Say:

- cr8 write 0
- cr8 write 5
- no exit to userspace
- userspace signals interrupt with priority 
4 because it knows about tpr == 0.