Hi Peter,
On 10/10/19 1:42 PM, Peter Zijlstra wrote:
On Thu, Oct 10, 2019 at 12:41:11PM +0200, Manfred Spraul wrote:
Hi,
Waiman Long noticed that the memory barriers in sem_lock() are not really
documented, and while adding documentation, I ended up with one case where
I'm not certain about the wake_q code:
Questions:
- Does smp_mb__before_atomic() + a (failed) cmpxchg_relaxed provide an
ordering guarantee?
Yep. Either the atomic instruction implies ordering (eg. x86 LOCK
prefix) or it doesn't (most RISC LL/SC), if it does,
smp_mb__{before,after}_atomic() are a NO-OP and the ordering is
unconditinoal, if it does not, then smp_mb__{before,after}_atomic() are
unconditional barriers.
And _relaxed() differs from "normal" cmpxchg only for LL/SC
architectures, correct?
Therefore smp_mb__{before,after}_atomic() may be combined with
cmpxchg_relaxed, to form a full memory barrier, on all archs.
[...]
- Is it ok that wake_up_q just writes wake_q->next, shouldn't
smp_store_acquire() be used? I.e.: guarantee that wake_up_process()
happens after cmpxchg_relaxed(), assuming that a failed cmpxchg_relaxed
provides any ordering.
There is no such thing as store_acquire, it is either load_acquire or
store_release. But just like how we can write load-aquire like
load+smp_mb(), so too I suppose we could write store-acquire like
store+smp_mb(), and that is exactly what is there (through the implied
barrier of wake_up_process()).
Thanks for confirming my assumption:
The code is correct, due to the implied barrier inside wake_up_process().
[...]
rewritten:
start condition: A = 1; B = 0;
CPU1:
B = 1;
RELEASE, unlock LockX;
CPU2:
lock LockX, ACQUIRE
if (LOAD A == 1) return; /* using cmp_xchg_relaxed */
CPU2:
A = 0;
ACQUIRE, lock LockY
smp_mb__after_spinlock();
READ B
Question: is A = 1, B = 0 possible?
Your example is incomplete (there is no A=1 assignment for example), but
I'm thinking I can guess where that should go given the earlier text.
A=1 is listed as start condition. Way before, someone did wake_q_add().
I don't think this is broken.
Thanks.
--
Manfred
