On 14/11/2018 20:42, Emilio G. Cota wrote:
> On Wed, Nov 14, 2018 at 12:44:00 +0100, Paolo Bonzini wrote:
>> This avoids the following deadlock:
>>
>> 1) a thread calls run_on_cpu for CPU 2 from a timer, and single_tcg_halt_cond
>> is signaled
>>
>> 2) CPU 1 is running and exits. It finds no work item and enters CPU 2
>>
>> 3) because the I/O thread is stuck in run_on_cpu, the round-robin kick
>> timer never triggers, and CPU 2 never runs the work item
>>
>> 4) run_on_cpu never completes
>
> I'm having trouble understanding (2)->(3).
>
> When the vCPU thread enters CPU 2, shouldn't it detect that work is
> pending? As in:
>
> /* assume cpu == cpu2 in the example above */
> while (cpu && !cpu->queued_work_first && !cpu->exit_request) {
>
> Both cpu->queued_work_first and cpu->exit_request will be set for cpu2.
>
> I can see though how with an additional CPU the deadlock
> could happen. For example, the I/O thread does run_on_cpu(cpu3),
> which kicks cpu1 (i.e. the tcg_current_rr_cpu) and cpu3, but not cpu2.
> Then cpu1 exits, and cpu2 starts executing; unless cpu2 exits on its
> own volition, it will run forever.
Yes, the thread must call run_on_cpu for CPU *3* from a timer.
Paolo
