[PATCH v2] nvme-multipath: Early exit if no path is available

[Hannes Reinecke]

On 1/29/21 8:45 AM, Chao Leng wrote:
> 
> 
> On 2021/1/29 15:06, Hannes Reinecke wrote:
>> On 1/29/21 4:07 AM, Chao Leng wrote:
>>>
>>>
>>> On 2021/1/29 9:42, Sagi Grimberg wrote:
>>>>
>>>>>> You can't see exactly where it dies but I followed the assembly to
>>>>>> nvme_round_robin_path(). Maybe it's not the initial 
>>>>>> nvme_next_ns(head,
>>>>>> old) which returns NULL but nvme_next_ns() is returning NULL 
>>>>>> eventually
>>>>>> (list_next_or_null_rcu()).
>>>>> So there is other bug cause nvme_next_ns abormal.
>>>>> I review the code about head->list and head->current_path, I find 2 
>>>>> bugs
>>>>> may cause the bug:
>>>>> First, I already send the patch. see:
>>>>> https://lore.kernel.org/linux-nvme/20210128033351.22116-1-lengchao@huawei.com/ 
>>>>>
>>>>> Second, in nvme_ns_remove, list_del_rcu is before
>>>>> nvme_mpath_clear_current_path. This may cause "old" is deleted from 
>>>>> the
>>>>> "head", but still use "old". I'm not sure there's any other
>>>>> consideration here, I will check it and try to fix it.
>>>>
>>>> The reason why we first remove from head->list and only then clear
>>>> current_path is because the other way around there is no way
>>>> to guarantee that that the ns won't be assigned as current_path
>>>> again (because it is in head->list).
>>> ok, I see.
>>>>
>>>> nvme_ns_remove fences continue of deletion of the ns by synchronizing
>>>> the srcu such that for sure the current_path clearance is visible.
>>> The list will be like this:
>>> head->next = ns1;
>>> ns1->next = head;
>>> old->next = ns1;
>>
>> Where does 'old' pointing to?
>>
>>> This may cause infinite loop in nvme_round_robin_path.
>>> for (ns = nvme_next_ns(head, old);
>>>      ns != old;
>>>      ns = nvme_next_ns(head, ns))
>>> The ns will always be ns1, and then infinite loop.
>>
>> No. nvme_next_ns() will return NULL.
> If there is just one path(the "old") and the "old" is deleted,
> nvme_next_ns() will return NULL.
> The list like this:
> head->next = head;
> old->next = head;
> If there is two or more path and the "old" is deleted,
> "for" will be infinite loop. because nvme_next_ns() will return
> the path which in the list except the "old", check condition will
> be true for ever.

But that will be caught by the statement above:

if (list_is_singular(&head->list))

no?

Cheers,

Hannes
-- 
Dr. Hannes Reinecke                Kernel Storage Architect
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