Gregory Haskins wrote:
> Avi Kivity wrote:
>> On 10/06/2009 04:22 PM, Gregory Haskins wrote:
>>>>>>>> +
>>>>>>>> +static inline void
>>>>>>>> +_kvm_xinterface_release(struct kref *kref)
>>>>>>>> +{
>>>>>>>> + struct kvm_xinterface *intf;
>>>>>>>> + struct module *owner;
>>>>>>>> +
>>>>>>>> + intf = container_of(kref, struct kvm_xinterface, kref);
>>>>>>>> +
>>>>>>>> + owner = intf->owner;
>>>>>>>> + rmb();
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>> Why rmb?
>>>>>>>
>>>>>>>
>>>>>> the intf->ops->release() line may invalidate the intf pointer, so we
>>>>>> want to ensure that the read completes before the release() is called.
>>>>>>
>>>>>> TBH: I'm not 100% its needed, but I was being conservative.
>>>>>>
>>>>>>
>>>>> rmb()s are only needed if an external agent can issue writes, otherwise
>>>>> you'd need one after every statement.
>>>>>
>>>> I was following lessons learned here:
>>>>
>>>> http://lkml.org/lkml/2009/7/7/175
>>>>
>>>> Perhaps mb() or barrier() are more appropriate than rmb()? I'm CC'ing
>>>> David Howells in case he has more insight.
>>>>
>>> BTW: In case it is not clear, the rationale as I understand it is we
>>> worry about the case where one cpu reorders the read to be after the
>>> ->release(), and another cpu might grab the memory that was kfree()'d
>>> within the ->release() and scribble something else on it before the read
>>> completes.
>>>
>>> I know rmb() typically needs to be paired with wmb() to be correct, so
>>> you are probably right to say that the rmb() itself is not appropriate.
>>> This problem in general makes my head hurt, which is why I said I am
>>> not 100% sure of what is required. As David mentions, perhaps
>>> "smp_mb()" is more appropriate for this application. I also speculate
>>> barrier() may be all that we need.
>>>
>> barrier() is the operation for a compiler barrier. But it's unneeded
>> here - unless the compiler can prove that ->release(intf) will not
>> modify intf->owner it is not allowed to move the access afterwards. An
>> indirect function call is generally a barrier() since the compiler can't
>> assume memory has not been modified.
>>
>
> You're logic
gak. or "your logic" even.
> seems reasonable to me. I will defer to David, since he
> brought up the issue with the similar logic originally.
>
> Kind Regards,
> -Greg
>
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