On Wed, Oct 2, 2013 at 12:19 PM, Waiman Long <waiman.long@xxxxxx> wrote:
> On 09/26/2013 06:42 PM, Jason Low wrote:
>>
>> On Thu, 2013-09-26 at 14:41 -0700, Tim Chen wrote:
>>>
>>> Okay, that would makes sense for consistency because we always
>>> first set node->lock = 0 at the top of the function.
>>>
>>> If we prefer to optimize this a bit though, perhaps we can
>>> first move the node->lock = 0 so that it gets executed after the
>>> "if (likely(prev == NULL)) {}" code block and then delete
>>> "node->lock = 1" inside the code block.
>>>
>>> static noinline
>>> void mcs_spin_lock(struct mcs_spin_node **lock, struct mcs_spin_node
>>> *node)
>>> {
>>> struct mcs_spin_node *prev;
>>>
>>> /* Init node */
>>> node->next = NULL;
>>>
>>> prev = xchg(lock, node);
>>> if (likely(prev == NULL)) {
>>> /* Lock acquired */
>>> return;
>>> }
>>> node->locked = 0;
>
>
> You can remove the locked flag setting statement inside if (prev == NULL),
> but you can't clear the locked flag after xchg(). In the interval between
> xchg() and locked=0, the previous lock owner may come in and set the flag.
> Now if your clear it, the thread will loop forever. You have to clear it
> before xchg().
Yes, in my most recent version, I left locked = 0 in its original
place so that the xchg() can act as a barrier for it.
The other option would have been to put another barrier after locked =
0. I went with leaving locked = 0 in its original place so that we
don't need that extra barrier.
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