Re: [patch -rt 1/2] KVM: use simple waitqueue for vcpu->wq

Paul Mackerras
2015-01-20T05:46:53Z

On Mon, Jan 19, 2015 at 12:41:00PM -0200, Marcelo Tosatti wrote:
> On Fri, Jan 16, 2015 at 11:48:46AM -0500, Steven Rostedt wrote:
> > > static void kvmppc_vcore_blocked(struct kvmppc_vcore *vc)
> > > {
> > > - DEFINE_WAIT(wait);
> > > + DEFINE_SWAITER(wait);
> > >
> > > - prepare_to_wait(&vc->wq, &wait, TASK_INTERRUPTIBLE);
> > > + swait_prepare(&vc->wq, &wait, TASK_INTERRUPTIBLE);
> > > vc->vcore_state = VCORE_SLEEPING;
> > > spin_unlock(&vc->lock);
> > > schedule();
> > > - finish_wait(&vc->wq, &wait);
> > > + swait_finish(&vc->wq, &wait);
> > > spin_lock(&vc->lock);
> > > vc->vcore_state = VCORE_INACTIVE;
> > > }
> > > @@ -1613,7 +1613,7 @@
> > > kvmppc_create_dtl_entry(vcpu, vc);
> > > kvmppc_start_thread(vcpu);
> > > } else if (vc->vcore_state == VCORE_SLEEPING) {
> > > - wake_up(&vc->wq);
> > > + swait_wake(&vc->wq);
> >
> > I notice everywhere you have a swait_wake_interruptible() but here. Is
> > there a reason why?
> >
> > IIRC, Peter wants to make swait wakeup usage homogenous. That is, you
> > either sleep in an interruptible state, or you don't. You can't mix and
> > match it.
>
> IIUC there is only one waiter on this waitqueue at any given time.
>
> Paul is that correct?

Yes, that's right. It's only the task that has taken the
responsibility for running the virtual core that would be waiting on
that wait queue.

Paul.