Low-Lock Singletons In D

[Brad Roberts]

On 5/7/13 3:44 PM, Andrei Alexandrescu wrote:
> On 5/7/13 5:57 PM, Mehrdad wrote:
>> On Tuesday, 7 May 2013 at 19:49:30 UTC, Andrei Alexandrescu wrote:
>>> A memory barrier is not a one-way thing, i.e. not only the writer must
>>> do it. Any operation on shared memory is a handshake between the
>>> writer and the reader. If the reader doesn't do its bit, it can see
>>> the writes out of order no matter what the writer does.
>>>
>>> Andrei
>>
>>
>>
>> Andrew, I still don't understand:
>>
>> The writer is ensuring that writes to memory are happening _after_ the
>> object is initialized and _before_ the reference to the old object is
>> modified, via a memory barrier.
>
> The writer is only half of the equation. The reader has its own cache to
> worry about and its own loading order.
>
>> Unless you're claiming that a memory barrier _doesn't_ do what it's
>> supposed to (i.e., the memory module is executing writes out-of-order
>> even though the processor is issuing them in the correct order), there
>> is no way for _anyone_ to see a partially initialized object anywhere...
>
> I'm not claiming, I'm destroying :o). There is. I know it's confusing.
> You may want to peruse the reading materials linked by others.
>
>
> Andrei

(this might be a repeat, I've only skimmed this thread)

Section 8.2 of this doc is a good read:
   http://download.intel.com/products/processor/manual/325384.pdf

There's a couple allowed re-orderings that are not what most people expect.

Later,
Brad