Re: [Qemu-devel] [RFC] create a single workqueue for each vm to update vm irq routing table

[Avi Kivity <avi@cloudius-systems.com>]

On 11/26/2013 04:46 PM, Paolo Bonzini wrote:
> Il 26/11/2013 15:36, Avi Kivity ha scritto:
>>      No, this would be exactly the same code that is running now:
>>
>>              mutex_lock(&kvm->irq_lock);
>>              old = kvm->irq_routing;
>>              kvm_irq_routing_update(kvm, new);
>>              mutex_unlock(&kvm->irq_lock);
>>
>>              synchronize_rcu();
>>              kfree(old);
>>              return 0;
>>
>>      Except that the kfree would run in the call_rcu kernel thread instead of
>>      the vcpu thread.  But the vcpus already see the new routing table after
>>      the rcu_assign_pointer that is in kvm_irq_routing_update.
>>
>> I understood the proposal was also to eliminate the synchronize_rcu(),
>> so while new interrupts would see the new routing table, interrupts
>> already in flight could pick up the old one.
> Isn't that always the case with RCU?  (See my answer above: "the vcpus
> already see the new routing table after the rcu_assign_pointer that is
> in kvm_irq_routing_update").

With synchronize_rcu(), you have the additional guarantee that any 
parallel accesses to the old routing table have completed.  Since we 
also trigger the irq from rcu context, you know that after 
synchronize_rcu() you won't get any interrupts to the old destination 
(see kvm_set_irq_inatomic()).

It's another question whether the hardware provides the same guarantee.

> If you eliminate the synchronize_rcu, new interrupts would see the new
> routing table, while interrupts already in flight will get a dangling
> pointer.

Sure, if you drop the synchronize_rcu(), you have to add call_rcu().