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

[Hannes Reinecke]

On 1/29/21 9:46 AM, Chao Leng wrote:
> 
> 
> On 2021/1/29 16:33, Hannes Reinecke wrote:
>> 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?
> Two path just a sample example.
> If there is just two path, will enter it, may cause no path but there is
> actually one path. It is falsely assumed that the "old" must be not 
> deleted.
> If there is more than two path, will cause infinite loop.
So you mean we'll need something like this?

diff --git a/drivers/nvme/host/multipath.c b/drivers/nvme/host/multipath.c
index 71696819c228..8ffccaf9c19a 100644
--- a/drivers/nvme/host/multipath.c
+++ b/drivers/nvme/host/multipath.c
@@ -202,10 +202,12 @@ static struct nvme_ns *__nvme_find_path(struct 
nvme_ns_head *head, int node)
  static struct nvme_ns *nvme_next_ns(struct nvme_ns_head *head,
                 struct nvme_ns *ns)
  {
-       ns = list_next_or_null_rcu(&head->list, &ns->siblings, struct 
nvme_ns,
-                       siblings);
-       if (ns)
-               return ns;
+       if (ns) {
+               ns = list_next_or_null_rcu(&head->list, &ns->siblings,
+                                          struct nvme_ns, siblings);
+               if (ns)
+                       return ns;
+       }
         return list_first_or_null_rcu(&head->list, struct nvme_ns, 
siblings);
  }

Cheers,

Hannes
-- 
Dr. Hannes Reinecke                Kernel Storage Architect
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