Subject: Re: [PATCH tip/locking/core v4 1/6] powerpc: atomic: Make *xchg and *cmpxchg a full barrier

From: Will Deacon <[email-masked]>
Date: Thu, 15 Oct 2015 11:35:44 +0100

Dammit guys, it's never simple is it?

On Wed, Oct 14, 2015 at 02:44:53PM -0700, Paul E. McKenney wrote:
> To that end, the herd tool can make a diagram of what it thought
> happened, and I have attached it.  I used this diagram to try and force
> this scenario at https://www.cl.cam.ac.uk/~pes20/ppcmem/index.html#PPC,
> and succeeded.  Here is the sequence of events:
> 
> o	Commit P0's write.  The model offers to propagate this write
> 	to the coherence point and to P1, but don't do so yet.
> 
> o	Commit P1's write.  Similar offers, but don't take them up yet.
> 
> o	Commit P0's lwsync.
> 
> o	Execute P0's lwarx, which reads a=0.  Then commit it.
> 
> o	Commit P0's stwcx. as successful.  This stores a=1.

On arm64, this is a conditional-store-*release* and therefore cannot be
observed before the initial write to x...

> o	Commit P0's branch (not taken).
> 
> o	Commit P0's final register-to-register move.
> 
> o	Commit P1's sync instruction.
> 
> o	There is now nothing that can happen in either processor.
> 	P0 is done, and P1 is waiting for its sync.  Therefore,
> 	propagate P1's a=2 write to the coherence point and to
> 	the other thread.

... therefore this is illegal, because you haven't yet propagated that
prior write...

> 
> o	There is still nothing that can happen in either processor.
> 	So pick the barrier propagate, then the acknowledge sync.
> 
> o	P1 can now execute its read from x.  Because P0's write to
> 	x is still waiting to propagate to P1, this still reads
> 	x=0.  Execute and commit, and we now have both r3 registers
> 	equal to zero and the final value a=2.

... and P1 would have to read x == 1.

So arm64 is ok. Doesn't lwsync order store->store observability for PPC?

Will