Re: [PATCH RFC 1/3] mutex: Make more scalable by doing less atomic operations

[Waiman Long <Waiman.Long@hp.com>]

On 04/08/2013 10:38 AM, Linus Torvalds wrote:
> On Mon, Apr 8, 2013 at 5:42 AM, Ingo Molnar<mingo@kernel.org>  wrote:
>> AFAICS the main performance trade-off is the following: when the owner CPU unlocks
>> the mutex, we'll poll it via a read first, which turns the cacheline into
>> shared-read MESI state. Then we notice that its content signals 'lock is
>> available', and we attempt the trylock again.
>>
>> This increases lock latency in the few-contended-tasks case slightly - and we'd
>> like to know by precisely how much, not just for a generic '10-100 users' case
>> which does not tell much about the contention level.
> We had this problem for *some* lock where we used a "read + cmpxchg"
> in the hotpath and it caused us problems due to two cacheline state
> transitions (first to shared, then to exclusive). It was faster to
> just assume it was unlocked and try to do an immediate cmpxchg.
>
> But iirc it is a non-issue for this case, because this is only about
> the contended slow path.
>
> I forget where we saw the case where we should *not* read the initial
> value, though. Anybody remember?
>
> That said, the MUTEX_SHOULD_XCHG_COUNT macro should die. Why shouldn't
> all architectures just consider negative counts to be locked? It
> doesn't matter that some might only ever see -1.

I think so too. However, I don't have the machines to test out other 
architectures. The MUTEX_SHOULD_XCHG_COUNT is just a safety measure to 
make sure that my code won't screw up the kernel in other architectures. 
Once it is confirmed that a negative count other than -1 is fine for all 
the other architectures, the macro can certainly go.

Regards,
Longman