That's what I meant in my first mail. We need to say whether this example will work or not. A=10 barrier barrier x = A If not, which synchronization functions need to be used and how? The current text does not automatically cover this example. Rajeev On Mar 15, 2011, at 9:28 PM, Torsten Hoefler wrote:

On Tue, Mar 15, 2011 at 09:22:49PM -0500, Rajeev Thakur wrote:

...

All can be used, but none are needed, right? We might need to talk about synch or flush. Some shared memory systems are loosely coupled and need membars to guarantee correct ordering.

...

Currently, in the unified model you can do:

lock put(A) unlock barrier barrier x = A

With shared memory, you could do

A=10 barrier barrier x = A

Right? Not really. The barrier and the x=A might be re-ordered. This is also the biggest issue with our definition of the unified model, some shared memory systems might actually not be able to support it :-(. We had a long discussion in Stuttgart about this.

...

If the above is true, it might be a bit confusing to users that in the RMA chapter you use put, get, and synchronization functions for some things and can directly read/write memory for some other things. Well, the standard is not a user documentation :-). But yes, having this special kind of window requires some effort to understand it.

Best, Torsten

-- bash$ :(){ :|:&};: --------------------- http://www.unixer.de/ ----- Torsten Hoefler | Performance Modeling and Simulation Lead Blue Waters Directorate | University of Illinois (UIUC) 1205 W Clark Street | Urbana, IL, 61801 NCSA Building | +01 (217) 244-7736 _______________________________________________ mpi3-rma mailing list [email protected] http://lists.mpi-forum.org/mailman/listinfo.cgi/mpi3-rma