linux-kernel - Re: Uses for memory barriers

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

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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