Date: Fri, 8 Sep 2006 14:06:29 -0400 (EDT)
From: Alan Stern <stern@rowland.harvard.edu>
To: Oliver Neukum <oliver@neukum.org>
cc: "Paul E. McKenney" <paulmck@us.ibm.com>,
       David Howells <dhowells@redhat.com>,
       Kernel development list <linux-kernel@vger.kernel.org>
Subject: Re: Uses for memory barriers
In-Reply-To: <200609081929.33027.oliver@neukum.org>
Message-ID: <Pine.LNX.4.44L0.0609081401270.7953-100000@iolanthe.rowland.org>
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On Fri, 8 Sep 2006, Oliver Neukum wrote:

> Am Freitag, 8. September 2006 17:55 schrieb Alan Stern:
> > > ????CPU 0???????????????????CPU 1
> > > ????-----???????????????????-----
> > > ????while (y==0) relax();???y = -1;
> > > ????a = 1;??????????????????b = 1;
> > > ????mb();???????????????????mb();
> > > ????y = b;??????????????????x = a;
> > > ????????????????????????????while (y < 0) relax();
> > > ????????????????????????????assert(x==1 || y==1);???//???
> 
> 					y =  -1
> 	a = 1
> 	y =b (== 0)
> 					b = 1
> 					x = a (==1)

So in this case the assertion is satisfied because x == 1.  My question
was whether the assertion could ever fail.  Paul said that it could, but I
didn't find his reason convincing.

> > Disagree: CPU 0 executes mb() between reading y and b. ?You have assumed
> > that CPU 1 executed its write to b and its mb() before CPU 0 got started, 
> > so why wouldn't CPU 0's mb() guarantee that it sees the new value of b? ?
> > That's really the key point.
> 
> That code has an ordinary race condition.
> For it to work it needs to be:
> 
> b = 1;
> mb(); //could be wmb()
> y = -1;

I'm not sure what you mean.  The code wasn't intended to "work" in any 
sense; it was just to make a point.  My question still stands: Is it 
possible, in the code as I originally wrote it, for the assertion to fail?

Alan Stern

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