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

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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.
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