Merge branch 'core-rcu-for-linus' of git://git.kernel.org/pub/scm/linux/kernel/git/tip/linux-2.6-tip

* 'core-rcu-for-linus' of git://git.kernel.org/pub/scm/linux/kernel/git/tip/linux-2.6-tip:
  smp: Document transitivity for memory barriers.
  rcu: add comment saying why DEBUG_OBJECTS_RCU_HEAD depends on PREEMPT.
  rcupdate: remove dead code
  rcu: add documentation saying which RCU flavor to choose
  rcutorture: Get rid of duplicate sched.h include
  rcu: call __rcu_read_unlock() in exit_rcu for tiny RCU

2 files changed, 89 insertions, 0 deletions

diff --git a/Documentation/RCU/whatisRCU.txt b/Documentation/RCU/whatisRCU.txt
index cfaac34c4557..6ef692667e2f 100644
--- a/Documentation/RCU/whatisRCU.txt
+++ b/Documentation/RCU/whatisRCU.txt
@@ -849,6 +849,37 @@ All:  lockdep-checked RCU-protected pointer access
 See the comment headers in the source code (or the docbook generated
 from them) for more information.
 
+However, given that there are no fewer than four families of RCU APIs
+in the Linux kernel, how do you choose which one to use?  The following
+list can be helpful:
+
+a.	Will readers need to block?  If so, you need SRCU.
+
+b.	What about the -rt patchset?  If readers would need to block
+	in an non-rt kernel, you need SRCU.  If readers would block
+	in a -rt kernel, but not in a non-rt kernel, SRCU is not
+	necessary.
+
+c.	Do you need to treat NMI handlers, hardirq handlers,
+	and code segments with preemption disabled (whether
+	via preempt_disable(), local_irq_save(), local_bh_disable(),
+	or some other mechanism) as if they were explicit RCU readers?
+	If so, you need RCU-sched.
+
+d.	Do you need RCU grace periods to complete even in the face
+	of softirq monopolization of one or more of the CPUs?  For
+	example, is your code subject to network-based denial-of-service
+	attacks?  If so, you need RCU-bh.
+
+e.	Is your workload too update-intensive for normal use of
+	RCU, but inappropriate for other synchronization mechanisms?
+	If so, consider SLAB_DESTROY_BY_RCU.  But please be careful!
+
+f.	Otherwise, use RCU.
+
+Of course, this all assumes that you have determined that RCU is in fact
+the right tool for your job.
+
 
 8.  ANSWERS TO QUICK QUIZZES
 
diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
index 631ad2f1b229..f0d3a8026a56 100644
--- a/Documentation/memory-barriers.txt
+++ b/Documentation/memory-barriers.txt
@@ -21,6 +21,7 @@ Contents:
      - SMP barrier pairing.
      - Examples of memory barrier sequences.
      - Read memory barriers vs load speculation.
+     - Transitivity
 
  (*) Explicit kernel barriers.
 
@@ -959,6 +960,63 @@ the speculation will be cancelled and the value reloaded:
 	retrieved                               :       :       +-------+
 
 
+TRANSITIVITY
+------------
+
+Transitivity is a deeply intuitive notion about ordering that is not
+always provided by real computer systems.  The following example
+demonstrates transitivity (also called "cumulativity"):
+
+	CPU 1			CPU 2			CPU 3
+	=======================	=======================	=======================
+		{ X = 0, Y = 0 }
+	STORE X=1		LOAD X			STORE Y=1
+				<general barrier>	<general barrier>
+				LOAD Y			LOAD X
+
+Suppose that CPU 2's load from X returns 1 and its load from Y returns 0.
+This indicates that CPU 2's load from X in some sense follows CPU 1's
+store to X and that CPU 2's load from Y in some sense preceded CPU 3's
+store to Y.  The question is then "Can CPU 3's load from X return 0?"
+
+Because CPU 2's load from X in some sense came after CPU 1's store, it
+is natural to expect that CPU 3's load from X must therefore return 1.
+This expectation is an example of transitivity: if a load executing on
+CPU A follows a load from the same variable executing on CPU B, then
+CPU A's load must either return the same value that CPU B's load did,
+or must return some later value.
+
+In the Linux kernel, use of general memory barriers guarantees
+transitivity.  Therefore, in the above example, if CPU 2's load from X
+returns 1 and its load from Y returns 0, then CPU 3's load from X must
+also return 1.
+
+However, transitivity is -not- guaranteed for read or write barriers.
+For example, suppose that CPU 2's general barrier in the above example
+is changed to a read barrier as shown below:
+
+	CPU 1			CPU 2			CPU 3
+	=======================	=======================	=======================
+		{ X = 0, Y = 0 }
+	STORE X=1		LOAD X			STORE Y=1
+				<read barrier>		<general barrier>
+				LOAD Y			LOAD X
+
+This substitution destroys transitivity: in this example, it is perfectly
+legal for CPU 2's load from X to return 1, its load from Y to return 0,
+and CPU 3's load from X to return 0.
+
+The key point is that although CPU 2's read barrier orders its pair
+of loads, it does not guarantee to order CPU 1's store.  Therefore, if
+this example runs on a system where CPUs 1 and 2 share a store buffer
+or a level of cache, CPU 2 might have early access to CPU 1's writes.
+General barriers are therefore required to ensure that all CPUs agree
+on the combined order of CPU 1's and CPU 2's accesses.
+
+To reiterate, if your code requires transitivity, use general barriers
+throughout.
+
+
 ========================
 EXPLICIT KERNEL BARRIERS
 ========================