Re: [musl] Illegal killlock skipping when transitioning to single-threaded state

[Szabolcs Nagy <nsz@port70.net>]

* Szabolcs Nagy <nsz@port70.net> [2022-10-03 15:26:15 +0200]:

> * Alexey Izbyshev <izbyshev@ispras.ru> [2022-10-03 09:16:03 +0300]:
> > On 2022-09-19 18:29, Rich Felker wrote:
> > > On Wed, Sep 07, 2022 at 03:46:53AM +0300, Alexey Izbyshev wrote:
> ...
> > > > Reordering the "libc.need_locks = -1" assignment and
> > > > UNLOCK(E->killlock) and providing a store barrier between them
> > > > should fix the issue.
> > > 
> > > I think this all sounds correct. I'm not sure what you mean by a store
> > > barrier between them, since all lock and unlock operations are already
> > > full barriers.
> > > 
> > 
> > Before sending the report I tried to infer the intended ordering semantics
> > of LOCK/UNLOCK by looking at their implementations. For AArch64, I didn't
> > see why they would provide a full barrier (my reasoning is below), so I
> > concluded that probably acquire/release semantics was intended in general
> > and suggested an extra store barrier to prevent hoisting of "libc.need_locks
> > = -1" store spelled after UNLOCK(E->killlock) back into the critical
> > section.
> > 
> > UNLOCK is implemented via a_fetch_add(). On AArch64, it is a simple
> > a_ll()/a_sc() loop without extra barriers, and a_ll()/a_sc() are implemented
> > via load-acquire/store-release instructions. Therefore, if we consider a
> > LOCK/UNLOCK critical section containing only plain loads and stores, (a) any
> > such memory access can be reordered with the initial ldaxr in UNLOCK, and
> > (b) any plain load following UNLOCK can be reordered with stlxr (assuming
> > the processor predicts that stlxr succeeds), and further, due to (a), with
> > any memory access inside the critical section. Therefore, UNLOCK is not full
> > barrier. Is this right?
> 
> i dont think this is right.


i think i was wrong and you are right.

so with your suggested swap of UNLOCK(killlock) and need_locks=-1 and
starting with 'something == 0' the exiting E and remaining R threads:

E:something=1      // protected by killlock
E:UNLOCK(killlock)
E:need_locks=-1

R:LOCK(unrelated)  // reads need_locks == -1
R:need_locks=0
R:UNLOCK(unrelated)
R:LOCK(killlock)   // does not lock
R:read something   // can it be 0 ?

and here something can be 0 (ie. not protected by killlock) on aarch64
because

T1
	something=1
	ldaxr ... killlock
	stlxr ... killlock
	need_locks=-1

T2
	x=need_locks
	ldaxr ... unrelated
	stlxr ... unrelated
	y=something

can end with x==-1 and y==0.

and to fix it, both a_fetch_add and a_cas need an a_barrier.

i need to think how to support such lock usage on aarch64
without adding too many dmb.



> 
> memory operations in the critical section cannot visibly happen after the
> final stlxr.
> 
> memory operations after the critical section cannot visibly happen before
> the ldaxr of UNLOCK.
> 
> the only issue can be inside the ll/sc loop in UNLOCK if there are independent
> memory operations there, but there arent.
> 
> > 
> > As for a store following UNLOCK, I'm not sure. A plain store following stlxr
> > can be reordered with it, but here that store is conditional on stlxr
> > success. So even if the processor predicts stlxr success, it can't make the
> > following store visible before it's sure that stlxr succeeded. But I don't
> > know whether the events "the processor knows that stlxr succeeded" and "the
> > result of stlxr is globally visible" are separate or not, and if they are
> > separate, what comes first. Depending on the answer, UNLOCK acts as a store
> > barrier or not.
> > 
> 
> UNLOCK on aarch64 acts as a full seqcst barrier as far as i can see.
> 
> but looking into the arch implementations needs a detailed understanding
> of the arch memory model (eg aarch64 stlxr is RCsc not RCpc like iso c
> release store), but there is no need for that: the musl model is
> simple seqcst synchronization everywhere.