Merge branch 'locking-urgent-for-linus' of git://git.kernel.org/pub/scm/linux/kernel/git/tip/tip

Pull locking fixes from Ingo Molnar:
 "Documentation updates and a bitops ordering fix"

* 'locking-urgent-for-linus' of git://git.kernel.org/pub/scm/linux/kernel/git/tip/tip:
  bitops: Do not default to __clear_bit() for __clear_bit_unlock()
  documentation: Clarify compiler store-fusion example
  documentation: Transitivity is not cumulativity
  documentation:  Add alternative release-acquire outcome
  documentation: Distinguish between local and global transitivity
  documentation: Subsequent writes ordered by rcu_dereference()
  documentation: Remove obsolete reference to RCU-protected indexes
  documentation: Fix memory-barriers.txt section references
  documentation: Fix control dependency and identical stores

1 files changed, 116 insertions, 25 deletions

diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
index 904ee42d078e..3729cbe60e41 100644
--- a/Documentation/memory-barriers.txt
+++ b/Documentation/memory-barriers.txt
@@ -232,7 +232,7 @@ And there are a number of things that _must_ or _must_not_ be assumed:
      with memory references that are not protected by READ_ONCE() and
      WRITE_ONCE().  Without them, the compiler is within its rights to
      do all sorts of "creative" transformations, which are covered in
-     the Compiler Barrier section.
+     the COMPILER BARRIER section.
 
  (*) It _must_not_ be assumed that independent loads and stores will be issued
      in the order given.  This means that for:
@@ -555,6 +555,30 @@ between the address load and the data load:
 This enforces the occurrence of one of the two implications, and prevents the
 third possibility from arising.
 
+A data-dependency barrier must also order against dependent writes:
+
+	CPU 1		      CPU 2
+	===============	      ===============
+	{ A == 1, B == 2, C = 3, P == &A, Q == &C }
+	B = 4;
+	<write barrier>
+	WRITE_ONCE(P, &B);
+			      Q = READ_ONCE(P);
+			      <data dependency barrier>
+			      *Q = 5;
+
+The data-dependency barrier must order the read into Q with the store
+into *Q.  This prohibits this outcome:
+
+	(Q == B) && (B == 4)
+
+Please note that this pattern should be rare.  After all, the whole point
+of dependency ordering is to -prevent- writes to the data structure, along
+with the expensive cache misses associated with those writes.  This pattern
+can be used to record rare error conditions and the like, and the ordering
+prevents such records from being lost.
+
+
 [!] Note that this extremely counterintuitive situation arises most easily on
 machines with split caches, so that, for example, one cache bank processes
 even-numbered cache lines and the other bank processes odd-numbered cache
@@ -565,21 +589,6 @@ odd-numbered bank is idle, one can see the new value of the pointer P (&B),
 but the old value of the variable B (2).
 
 
-Another example of where data dependency barriers might be required is where a
-number is read from memory and then used to calculate the index for an array
-access:
-
-	CPU 1		      CPU 2
-	===============	      ===============
-	{ M[0] == 1, M[1] == 2, M[3] = 3, P == 0, Q == 3 }
-	M[1] = 4;
-	<write barrier>
-	WRITE_ONCE(P, 1);
-			      Q = READ_ONCE(P);
-			      <data dependency barrier>
-			      D = M[Q];
-
-
 The data dependency barrier is very important to the RCU system,
 for example.  See rcu_assign_pointer() and rcu_dereference() in
 include/linux/rcupdate.h.  This permits the current target of an RCU'd
@@ -800,9 +809,13 @@ In summary:
       use smp_rmb(), smp_wmb(), or, in the case of prior stores and
       later loads, smp_mb().
 
-  (*) If both legs of the "if" statement begin with identical stores
-      to the same variable, a barrier() statement is required at the
-      beginning of each leg of the "if" statement.
+  (*) If both legs of the "if" statement begin with identical stores to
+      the same variable, then those stores must be ordered, either by
+      preceding both of them with smp_mb() or by using smp_store_release()
+      to carry out the stores.  Please note that it is -not- sufficient
+      to use barrier() at beginning of each leg of the "if" statement,
+      as optimizing compilers do not necessarily respect barrier()
+      in this case.
 
   (*) Control dependencies require at least one run-time conditional
       between the prior load and the subsequent store, and this
@@ -814,7 +827,7 @@ In summary:
   (*) Control dependencies require that the compiler avoid reordering the
       dependency into nonexistence.  Careful use of READ_ONCE() or
       atomic{,64}_read() can help to preserve your control dependency.
-      Please see the Compiler Barrier section for more information.
+      Please see the COMPILER BARRIER section for more information.
 
   (*) Control dependencies pair normally with other types of barriers.
 
@@ -1257,7 +1270,7 @@ 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"):
+demonstrates transitivity:
 
 	CPU 1			CPU 2			CPU 3
 	=======================	=======================	=======================
@@ -1305,8 +1318,86 @@ 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.
+General barriers provide "global transitivity", so that all CPUs will
+agree on the order of operations.  In contrast, a chain of release-acquire
+pairs provides only "local transitivity", so that only those CPUs on
+the chain are guaranteed to agree on the combined order of the accesses.
+For example, switching to C code in deference to Herman Hollerith:
+
+	int u, v, x, y, z;
+
+	void cpu0(void)
+	{
+		r0 = smp_load_acquire(&x);
+		WRITE_ONCE(u, 1);
+		smp_store_release(&y, 1);
+	}
+
+	void cpu1(void)
+	{
+		r1 = smp_load_acquire(&y);
+		r4 = READ_ONCE(v);
+		r5 = READ_ONCE(u);
+		smp_store_release(&z, 1);
+	}
+
+	void cpu2(void)
+	{
+		r2 = smp_load_acquire(&z);
+		smp_store_release(&x, 1);
+	}
+
+	void cpu3(void)
+	{
+		WRITE_ONCE(v, 1);
+		smp_mb();
+		r3 = READ_ONCE(u);
+	}
+
+Because cpu0(), cpu1(), and cpu2() participate in a local transitive
+chain of smp_store_release()/smp_load_acquire() pairs, the following
+outcome is prohibited:
+
+	r0 == 1 && r1 == 1 && r2 == 1
+
+Furthermore, because of the release-acquire relationship between cpu0()
+and cpu1(), cpu1() must see cpu0()'s writes, so that the following
+outcome is prohibited:
+
+	r1 == 1 && r5 == 0
+
+However, the transitivity of release-acquire is local to the participating
+CPUs and does not apply to cpu3().  Therefore, the following outcome
+is possible:
+
+	r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0
+
+As an aside, the following outcome is also possible:
+
+	r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0 && r5 == 1
+
+Although cpu0(), cpu1(), and cpu2() will see their respective reads and
+writes in order, CPUs not involved in the release-acquire chain might
+well disagree on the order.  This disagreement stems from the fact that
+the weak memory-barrier instructions used to implement smp_load_acquire()
+and smp_store_release() are not required to order prior stores against
+subsequent loads in all cases.  This means that cpu3() can see cpu0()'s
+store to u as happening -after- cpu1()'s load from v, even though
+both cpu0() and cpu1() agree that these two operations occurred in the
+intended order.
+
+However, please keep in mind that smp_load_acquire() is not magic.
+In particular, it simply reads from its argument with ordering.  It does
+-not- ensure that any particular value will be read.  Therefore, the
+following outcome is possible:
+
+	r0 == 0 && r1 == 0 && r2 == 0 && r5 == 0
+
+Note that this outcome can happen even on a mythical sequentially
+consistent system where nothing is ever reordered.
+
+To reiterate, if your code requires global transitivity, use general
+barriers throughout.
 
 
 ========================
@@ -1459,7 +1550,7 @@ of optimizations:
      the following:
 
 	a = 0;
-	/* Code that does not store to variable a. */
+	... Code that does not store to variable a ...
 	a = 0;
 
      The compiler sees that the value of variable 'a' is already zero, so
@@ -1471,7 +1562,7 @@ of optimizations:
      wrong guess:
 
 	WRITE_ONCE(a, 0);
-	/* Code that does not store to variable a. */
+	... Code that does not store to variable a ...
 	WRITE_ONCE(a, 0);
 
  (*) The compiler is within its rights to reorder memory accesses unless