Re: Arm ldaxr / stxr loop question

On 11/8/24 17:56, Chris M. Thomasson wrote:

On 11/8/2024 2:45 PM, Scott Lurndal wrote:

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

On 11/2/2024 12:10 PM, Chris M. Thomasson wrote:

On 11/1/2024 9:17 AM, aph@littlepinkcloud.invalid wrote:

jseigh <jseigh_es00@xemaps.com> wrote:

So if were to implement a spinlock using the above instructions
something along the lines of

>

>
Fwiw, I am basically asking if the "store" stxr has implied acquire
semantics wrt the "load" ldaxr? I am guess that it does... This would
imply that the acquire membar (#LoadStore | #LoadLoad) would be
respected by the store at stxr wrt its "attached?" load wrt ldaxr?
>
Is this basically right? Or, what am I missing here? Thanks.
>
The membar logic wrt acquire needs to occur _after_ the atomic logic
that locks the spinlock. A release barrier (#LoadStore | #StoreStore)
needs to occur _before_ the atomic logic that unlocks said spinlock.
>
Am I missing anything wrt ARM? ;^o

>
Did you read the extensive description of memory semantics
in the ARMv8 ARM?   See page 275 in DDI0487K_a.
>
https://developer.arm.com/documentation/ddi0487/ka/?lang=en

 I did not! So I am flying a mostly blind here. I don't really have any experience with how ARM handles these types of things. Just guessing that the store would honor the acquire of the load? Or, does the store need a membar and the load does not need acquire at all? I know that the membar should be after the final store that actually locks the spinlock wrt Joe's example.
 I just need to RTFM!!!!
 Sorry about that Scott. ;^o
 Perhaps sometime tonight. Is seems like optimistic LL/SC instead of pessimistic CAS RMW type of logic?

In this case the the stxr doesn't need a memory barrier.
Loads can move forward of it but not forward of the ldaxr
because it has acquire semantics.  For a lock that's ok
since the stxr would fail if any other thread acquired
the lock the conditional branch would make the loads
speculative if the stxr failed I believe.
Joe Seigh