> Note that it says the store into the first operand will appear to occur > before the store into > the second operand, but it does NOT say that an observing CPU will see both > stores or neither.
Again, not arguing, just trying to fully understand. In your answer above, did you consider this sentence from the PoOp description? "A serialization function is performed before the operation begins and again after the operation is completed." Charles -----Original Message----- From: IBM Mainframe Discussion List [mailto:IBM-MAIN@LISTSERV.UA.EDU] On Behalf Of Walt Farrell Sent: Tuesday, September 12, 2017 5:08 PM To: IBM-MAIN@LISTSERV.UA.EDU Subject: Re: CSST question On Tue, 12 Sep 2017 16:36:44 +0200, Charles Mills <charl...@mcn.org> wrote: >Disabling for interruptions is not sufficient in a multi-processor world, >right? > >I don't pretend to be the world's biggest machine instruction expert. >Am I reading the PoOp correctly that a task wishing another task's CSST >to effectively appear to be entirely atomic (from its CPU's point of view) >could achieve that effect by issuing a serialization instruction (BCR 15,0)? I won't pretend to be an expert, either, but I do not think a serialization operation will make CSST appear fully atomic to other CPUs. Serialization ensures that all operand fetches by other CPUs, and all operand stores by other CPUs, that occurred conceptually before the serializing instruction, will complete before the serializing instruction resumes. ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
