Low-Lock Singletons In D

[Dmitry Olshansky]

07-May-2013 23:49, Andrei Alexandrescu пишет:
> On 5/7/13 12:46 PM, Steven Schveighoffer wrote:
>> On Tue, 07 May 2013 12:30:05 -0400, deadalnix <deadalnix at gmail.com>
>> wrote:
>>
[snip]
>>>
>>> That is incorrect as the thread not going into the lock can see a
>>> partially initialized object.
>>
>> The memory barrier prevents that. You don't store the variable until the
>> object is initialized. That is the whole point.
>
> 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.
>
Exactly.


Returning to the confusing point.

On x86 things are actually muddied by stronger then required hardware 
guarantees. And only because of this there no need for explicit read 
barrier (nor x86 have one) in this case. Still the read operation has to 
be marked specifically (volatile, asm block, whatever) to ensure the 
_compiler_ does the right thing (no reordering of across it).


-- 
Dmitry Olshansky