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SubjectRe: [PATCH RFC tip/core/rcu 06/11] smp: Document transitivity for memory barriers.
On 02/23/2011 11:29 AM, Steven Rostedt wrote:
> On Tue, 2011-02-22 at 17:39 -0800, Paul E. McKenney wrote:
>> Transitivity is guaranteed only for full memory barriers (smp_mb()).
>>
>> Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com>
>> ---
>> Documentation/memory-barriers.txt | 58 +++++++++++++++++++++++++++++++++++++
>> 1 files changed, 58 insertions(+), 0 deletions(-)
>>
>> diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
>> index 631ad2f..f0d3a80 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.
>
> Sounds like someone had a fun time debugging their code.
>
>> +
>> +To reiterate, if your code requires transitivity, use general barriers
>> +throughout.
>
> I expect that your code is the only code in the kernel that actually
> requires transitivity ;-)
>

Maybe, but my RCURING also requires transitivity, I had asked Paul for advice
one years ago when I was writing the patch. Good document for it!

Thanks,
Lai


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