On Thu, Oct 03, 2013 at 10:29:59PM -0700, Paul E. McKenney wrote: > On Thu, Oct 03, 2013 at 04:28:27PM -0700, Josh Triplett wrote: > > On Thu, Oct 03, 2013 at 01:52:45PM -0700, Linus Torvalds wrote: > > > On Thu, Oct 3, 2013 at 1:41 PM, Al Viro <viro@xxxxxxxxxxxxxxxxxx> wrote: > > > > > > > > The problem is this: > > > > A = 1, B = 1 > > > > CPU1: > > > > A = 0 > > > > <full barrier> > > > > synchronize_rcu() > > > > read B > > > > > > > > CPU2: > > > > rcu_read_lock() > > > > B = 0 > > > > read A > > /me scratches his head... > > OK, for CPU2 to see 1 from its read from A, the corresponding RCU > read-side critical section must have started before CPU1 did A=0. This > means that this same RCU read-side critical section must have started > before CPU1's synchronize_rcu(), which means that it must complete > before that synchronize_rcu() returns. Therefore, CPU2's B=0 must > execute before CPU1's read of B, hence that read of B must return zero. > > Conversely, if CPU1's read from B returns 1, we know that CPU2's > RCU read-side critical section must not have completed until after > CPU1's synchronize_rcu() returned, which means that the RCU read-side > critical section must have started after that synchronize_rcu() started, > so CPU1's assignment to A must also have already happened. Therefore, > CPU2's read from A must return zero. Yeah, that makes sense. I think too much time spent staring at the *implementation* of RCU and the exciting assumptions it has to make about barriers or memory operations leaking out of the implementations of the RCU primitives (for instance, the fun needed to guarantee a memory barrier on all CPUs, or to safely use non-atomic operations inside RCU itself) makes it entirely too difficult to look at a perfectly ordinary *use* of RCU primitives and see the obvious. :) - Josh Triplett -- To unsubscribe from this list: send the line "unsubscribe linux-fsdevel" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
