Re: [Qemu-devel] [PATCH v2 for-2.5] rcu: Allow calling rcu_(un)register_thread() during synchronize_rcu()

[Wen Congyang <ghostwcy@gmail.com>]

At 2015/7/28 19:46, Paolo Bonzini Wrote:
>
>
> On 28/07/2015 12:33, Wen Congyang wrote:
>> On 07/28/2015 06:18 PM, Paolo Bonzini wrote:
>>>
>>>
>>> On 28/07/2015 12:02, Wen Congyang wrote:
>>>> I have a question about rcu: while do we call wait_for_readers()
>>>> twice for 32-bit host?
>>>
>>> Because there is a very small but non-zero probability of the counter
>>> going up by exactly 2^31 periods (periods are stored in bits 1-31 so you
>>> lose one bit) while the thread is sleeping.  This detail of the
>>> implementation comes from URCU.
>>
>> Yes, so you use rcu_gp_ctr ^ RCU_GP_CTR to instead of rcu_gp_ctr + RCU_GP_CTR.
>> The initial value is 1, so rcu_gp_ctr is: 1, 3, 1, 3, ...
>> The rcu_gp_ctr will never be 0. I think calling wait_for_readers() once is
>> enough.
>>
>> Do I miss something?
>
> If you call it just once, you have the same problem as before.  In fact,
> it's worse because instead of having an overflow every 2^31 periods, you
> have one every 2 periods.  Instead, by checking that rcu_reader went
> through 1 _and_ 3 (or that it was at least once 0, i.e. the thread was
> quiescent), you are sure that the thread went through _at least one_
> grace period.

The overflow is acceptable. We only compare if rcu_reader.ctr is equal than
rcu_gp_ctr. If not, we should wait that thread to call rcu_read_unlock().
We don't care which is bigger. If no threads calls sync_rcu(), all threads
rcu_read.ctr is 0 or rcu_gp_ctr. If one thread calls sync_rcu(), all threads
rcu_read.ctr is 0, old_rcu_gp_ctr, or new_rcu_gp_ctr. We only wait the 
thread
that's rcu_read.ctr is old_rcu_gp_ctr.

Thanks
Wen Congyang

>
> Paolo
>
>