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
This commit is contained in:
Linus Torvalds 2011-03-16 08:10:07 -07:00
Родитель 34d211a2d5 241e6663b5
Коммит 016aa2ed1c
5 изменённых файлов: 95 добавлений и 7 удалений

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@ -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

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@ -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
========================

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@ -214,11 +214,12 @@ static int rcuhead_fixup_free(void *addr, enum debug_obj_state state)
* Ensure that queued callbacks are all executed.
* If we detect that we are nested in a RCU read-side critical
* section, we should simply fail, otherwise we would deadlock.
* Note that the machinery to reliably determine whether
* or not we are in an RCU read-side critical section
* exists only in the preemptible RCU implementations
* (TINY_PREEMPT_RCU and TREE_PREEMPT_RCU), which is why
* DEBUG_OBJECTS_RCU_HEAD is disallowed if !PREEMPT.
*/
#ifndef CONFIG_PREEMPT
WARN_ON(1);
return 0;
#else
if (rcu_preempt_depth() != 0 || preempt_count() != 0 ||
irqs_disabled()) {
WARN_ON(1);
@ -229,7 +230,6 @@ static int rcuhead_fixup_free(void *addr, enum debug_obj_state state)
rcu_barrier_bh();
debug_object_free(head, &rcuhead_debug_descr);
return 1;
#endif
default:
return 0;
}

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@ -852,7 +852,7 @@ void exit_rcu(void)
if (t->rcu_read_lock_nesting == 0)
return;
t->rcu_read_lock_nesting = 1;
rcu_read_unlock();
__rcu_read_unlock();
}
#else /* #ifdef CONFIG_TINY_PREEMPT_RCU */

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@ -47,7 +47,6 @@
#include <linux/srcu.h>
#include <linux/slab.h>
#include <asm/byteorder.h>
#include <linux/sched.h>
MODULE_LICENSE("GPL");
MODULE_AUTHOR("Paul E. McKenney <paulmck@us.ibm.com> and "