Subj : Re: [.NET] Volatile Fields [correction]
To   : comp.programming.threads
From : David Hopwood
Date : Tue Sep 20 2005 01:59 pm

Joe Seigh wrote:
> Chris Thomasson wrote:
> 
>> [...] This works in x86:
>>
>> Processor 1:  Y = 1; L = 1;
>> Processor 2:  if (L == 1) { Z = 1; L = 2; }
>> Processor 3:  if (L == 2) { assert( Z == 1 && Y == 1 ); }
>>
>> http://groups.google.com/group/comp.programming.threads/msg/68ba70e66d6b6ee9?hl=en
>>
>> L is a lightweight memory barrier.
>
> You don't need the load on the "memory barrier".
> 
> W = X = Y = Z = 0;
> 
> Processor 1:   W = 1; L = 1; X = 1
> Processor 2:   if (X == 1) { Y = 1; L = 1; Z = 1; }
> Processor 3:   if (Z == 1) { assert (W == 1 && Y == 1); }
> 
> W and Y are stored by two different processors and are seen in order.

What are the stores to L for?

  W = X = Y = Z = 0;

  Processor 1:   W = 1; X = 1;
  Processor 2:   if (X == 1) { Y = 1; Z = 1; }
  Processor 3:   if (Z == 1) { assert (W == 1 && Y == 1); }

cannot fail the assert (for processor consistency / x86 WB / .NET with volatiles).

-- 
David Hopwood <david.nospam.hopwood@blueyonder.co.uk>

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