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(). -- To unsubscribe from this list: send the line "unsubscribe kvm" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
