Il 26/11/2013 16:03, Gleb Natapov ha scritto:
>>>> > >>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()).
> We do not have this guaranty for other vcpus that do not call
> synchronize_rcu(). They may still use outdated routing table while a vcpu
> or iothread that performed table update sits in synchronize_rcu().
Avi's point is that, after the VCPU resumes execution, you know that no
interrupt will be sent to the old destination because
kvm_set_msi_inatomic (and ultimately kvm_irq_delivery_to_apic_fast) is
also called within the RCU read-side critical section.
Without synchronize_rcu you could have
VCPU writes to routing table
e = entry from IRQ routing table
kvm_irq_routing_update(kvm, new);
VCPU resumes execution
kvm_set_msi_irq(e, &irq);
kvm_irq_delivery_to_apic_fast();
where the entry is stale but the VCPU has already resumed execution.
If we want to ensure, we need to use a different mechanism for
synchronization than the global RCU. QRCU would work; readers are not
wait-free but only if there is a concurrent synchronize_qrcu, which
should be rare.
Paolo
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