Re: [PATCH v5 3/5] locking/qspinlock: Introduce CNA into the slow path of qspinlock

Alex Kogan
2019-10-18T21:39:13Z

> On Oct 18, 2019, at 12:03 PM, Waiman Long <longman@redhat.com> wrote:
>
> On 10/16/19 12:29 AM, Alex Kogan wrote:
>> +static inline void cna_pass_lock(struct mcs_spinlock *node,
>> + struct mcs_spinlock *next)
>> +{
>> + struct cna_node *cn = (struct cna_node *)node;
>> + struct mcs_spinlock *next_holder = next, *tail_2nd;
>> + u32 val = 1;
>> +
>> + u32 scan = cn->pre_scan_result;
>> +
>> + /*
>> + * check if a successor from the same numa node has not been found in
>> + * pre-scan, and if so, try to find it in post-scan starting from the
>> + * node where pre-scan stopped (stored in @pre_scan_result)
>> + */
>> + if (scan > 0)
>> + scan = cna_scan_main_queue(node, decode_tail(scan));
>> +
>> + if (!scan) { /* if found a successor from the same numa node */
>> + next_holder = node->next;
>> + /*
>> + * make sure @val gets 1 if current holder's @locked is 0 as
>> + * we have to store a non-zero value in successor's @locked
>> + * to pass the lock
>> + */
>> + val = node->locked + (node->locked == 0);
>
> node->locked can be 0 when the cpu enters into an empty MCS queue. We
> could unconditionally set node->locked to 1 for this case in qspinlock.c
> or with your above code.

Right, I was doing that in the first two versions of the series. It adds
unnecessary store into @locked for non-CNA variants, and even if it does not
have any real performance implications, I think Peter did not like that (or,
at least, the comment I had to explain why we needed that store).

> Perhaps, a comment about when node->locked will
> be 0.
Yeah, I was tinkering with this comment. Here is how it read in v3:
/*
* We unlock a successor by passing a non-zero value,
* so set @val to 1 iff @locked is 0, which will happen
* if we acquired the MCS lock when its queue was empty
*/

I can change back to something like that if it is better.

>
> It may be easier to understand if you just do
>
> val = node->locked ? node->locked : 1;
You’re right, that’s another possibility.
However, it adds yet another if-statement on the critical path, which I was
trying to avoid that.

Best regards,
— Alex