On Thu, 25 Jul 2024 at 22:17, Paolo Bonzini <pbonzini@xxxxxxxxxx> wrote:
>
> On Thu, Jul 25, 2024 at 4:00 PM Liam Ni <zhiguangni01@xxxxxxxxx> wrote:
> >
> > The input parameter level to the pic_irq_request function indicates
> > whether there are interrupts to be injected,
> > a level value of 1 indicates that there are interrupts to be injected,
> > and a level value of 0 indicates that there are no interrupts to be injected.
> > And the value of level will be assigned to s->output,
> > so we should set s->wakeup_needed to true when s->output is true.
> >
> > Signed-off-by: Liam Ni <zhiguangni01@xxxxxxxxx>
> > ---
> > arch/x86/kvm/i8259.c | 2 +-
> > 1 file changed, 1 insertion(+), 1 deletion(-)
> >
> > diff --git a/arch/x86/kvm/i8259.c b/arch/x86/kvm/i8259.c
> > index 8dec646e764b..ec9d6ee7d33d 100644
> > --- a/arch/x86/kvm/i8259.c
> > +++ b/arch/x86/kvm/i8259.c
> > @@ -567,7 +567,7 @@ static void pic_irq_request(struct kvm *kvm, int level)
> > {
> > struct kvm_pic *s = kvm->arch.vpic;
> >
> > - if (!s->output)
> > + if (s->output)
>
> This is the old value of s->output. wakeup is needed if you have a
> 0->1 transition, so what you're looking for is either
I would like to know the reason why we monitor the 0->1 transformations?
>
> if (level)
> s->wakeup_needed = true;
This solution seems more appropriate with level=true,
indicating that there is a pending interrupt in the PIC
Thanks
Liam Ni
>
> or
>
> if (!s->output && level)
> s->wakeup_needed = true;
>
> but your version is incorrect because it would look for a 1->1
> transition instead.
>
> Thanks,
>
> Paolo
>
