Re: [RFC] llist: Fix code comments about llist_del_first locking
From: Huang\, Ying
Date: Thu Dec 08 2016 - 21:26:48 EST
Joel Fernandes <joelaf@xxxxxxxxxx> writes:
> On Thu, Dec 8, 2016 at 6:12 PM, Huang, Ying <ying.huang@xxxxxxxxx> wrote:
>> Joel Fernandes <joelaf@xxxxxxxxxx> writes:
>>
>>> On Thu, Dec 8, 2016 at 4:42 PM, Joel Fernandes <joelaf@xxxxxxxxxx> wrote:
>>>> On Thu, Dec 8, 2016 at 4:35 PM, Huang, Ying <ying.huang@xxxxxxxxx> wrote:
>>>>> Joel Fernandes <joelaf@xxxxxxxxxx> writes:
>>>>>
>>>>>> Usage llist_del_first needs lock protection, however the table in the
>>>>>> comments of llist.h show a '-'. Correct this, and also add better
>>>>>> comments on top.
>>>>>>
>>>>>> Cc: Huang Ying <ying.huang@xxxxxxxxx>
>>>>>> Cc: Ingo Molnar <mingo@xxxxxxxxxx>
>>>>>> Cc: Will Deacon <will.deacon@xxxxxxx>
>>>>>> Cc: Paul McKenney <paulmck@xxxxxxxxxxxxxxxxxx>
>>>>>> Signed-off-by: Joel Fernandes <joelaf@xxxxxxxxxx>
>>>>>> ---
>>>>>> include/linux/llist.h | 19 ++++++++++---------
>>>>>> 1 file changed, 10 insertions(+), 9 deletions(-)
>>>>>>
>>>>>> diff --git a/include/linux/llist.h b/include/linux/llist.h
>>>>>> index fd4ca0b..15e4949 100644
>>>>>> --- a/include/linux/llist.h
>>>>>> +++ b/include/linux/llist.h
>>>>>> @@ -3,14 +3,15 @@
>>>>>> /*
>>>>>> * Lock-less NULL terminated single linked list
>>>>>> *
>>>>>> - * If there are multiple producers and multiple consumers, llist_add
>>>>>> - * can be used in producers and llist_del_all can be used in
>>>>>> - * consumers. They can work simultaneously without lock. But
>>>>>> - * llist_del_first can not be used here. Because llist_del_first
>>>>>> - * depends on list->first->next does not changed if list->first is not
>>>>>> - * changed during its operation, but llist_del_first, llist_add,
>>>>>> - * llist_add (or llist_del_all, llist_add, llist_add) sequence in
>>>>>> - * another consumer may violate that.
>>>>>> + * If there are multiple producers and multiple consumers, llist_add can be
>>>>>> + * used in producers and llist_del_all can be used in consumers. They can work
>>>>>> + * simultaneously without lock. But llist_del_first will need to use a lock
>>>>>> + * with any other operation (ABA problem). This is because llist_del_first
>>>>>> + * depends on list->first->next not changing but there's no way to be sure
>>>>>> + * about that and the cmpxchg in llist_del_first may succeed if list->first is
>>>>>> + * the same after concurrent operations. For example, a llist_del_first,
>>>>>> + * llist_add, llist_add (or llist_del_all, llist_add, llist_add) sequence in
>>>>>> + * another consumer may cause violations.
>>>>>> *
>>>>>> * If there are multiple producers and one consumer, llist_add can be
>>>>>> * used in producers and llist_del_all or llist_del_first can be used
>>>>>> @@ -19,7 +20,7 @@
>>>>>> * This can be summarized as follow:
>>>>>> *
>>>>>> * | add | del_first | del_all
>>>>>> - * add | - | - | -
>>>>>> + * add | - | L | -
>>>>>
>>>>> If there are only one consumer which only calls llist_del_first(), lock
>>>>> is unnecessary. So '-' is shown here originally. But if there are
>>>>> multiple consumers which call llist_del_first() or llist_del_all(), lock
>>>>> is needed.
>>>>
>>>> I think this needs to be made more clear in the table. The table
>>>> doesn't clear say whether it describes the preceding paragraph
>>>> (multiple producers and one consumer), or if it describes the multiple
>>>> producers and one consumer case. So either we should have 2 tables, or
>>>
>>> Sorry, I meant "or if it describes the multiple producer and multiple
>>> consumer case".
>>
>> I tried to describe both cases in the original table.
>>
>> * | add | del_first | del_all
>> * add | - | - | -
>> * del_first | | L | L
>> * del_all | | | -
>>
>> The 'L' for "del_first * del_first" means multiple consumers uses
>> llist_del_first() need lock. And the 'L' for 'del_first * del_all'
>> means multiple consumers uses llist_del_first() and llist_del_all() need
>> lock.
>
> Ok, now I get it - so basically the table describes one
> producer/consumer vs another producer/consumer, in other words you are
> just describing contention between any 2 operations. Thanks for
> clarifying. I will respin the comments to explain this a bit better if
> that's Ok with you.
It is good for me to improve the comments.
Best Regards,
Huang, Ying