[concurrency-interest] Enforcing total sync order on modern hardware

Oleksandr Otenko oleksandr.otenko at oracle.com
Tue Mar 24 10:32:07 EDT 2015


On 24/03/2015 12:54, Marko Topolnik wrote:
> On Tue, Mar 24, 2015 at 1:47 PM, Oleksandr Otenko 
> <oleksandr.otenko at oracle.com <mailto:oleksandr.otenko at oracle.com>> wrote:
>
>     I don't know why you call IRIW a test of TSO, if total order of
>     reads is also important. Don't forget the load-load barrier, which
>     is implicit on x86, and a small theorem about what happens when
>     you don't order concurrent reads totally.
>
>
> But note that you still don't need a total _global_ order of reads for 
> IRIW, it just has to be preserved locally for each processor.
Some, but not all, reorderings of concurrent reads are indistinguishable 
from a certain fixed global total order of reads. What does this give 
you apart from freedom of implementation?


> Dekker's idiom exercises this more strongly because there has to be 
> global ordering between stores and loads by different processors.

Global ordering in Dekker's idiom you are talking about is also present 
in IRIW. Dekker's idiom specifies more in program order:

T1:
x=1;
r0=x;  // program order po(r0=x, x=1); in IRIW it is in synchronization 
order, so may be so(r0=x, x=1) or the other way around
r1=y;  // "global order" between r1=y and y=1 executed by different 
processors in Dekker's idiom is also present in IRIW

T2:
y=1;
r2=y;
r3=x;

The outcome r1==0, r3==0 is disallowed through the same reasoning in 
Dekker's idiom as in IRIW: since r1==0, y=1 must not happen before r1=y; 
then r3=x, being after y=1 in program order, must also not precede r1=y, 
and transitively cannot precede x=1. You need this step that from "Y 
must not precede X" follows that "X happens-before Y" - but that's what 
makes the order total.

In IRIW you have the same chain of reasoning. Even if you get rid of 
synchronization order, but keep synchronizes-with and transitive 
closure, then from observing r0==1 and r2==1 you still have the same 
edges. The key here still is the conclusion that since r1==0, y=1 must 
not happen before r1=y, which is the only part that requires total 
ordering - cannot be partially ordered. But it is exactly the same bit 
that cannot be partially ordered in Dekker's idiom: r1=y happens-before 
y=1 /because/ */otherwise/* r1=y /should/ observe 1. If you don't have 
mutual exclusion in ordering r1=y and y=1 in Dekker's idiom, why would 
the outcome r1==0, r3==0 be forbidden?


I've seen others criticizing IRIW as being not realistic, but can you 
support Dekker's idiom (which I reckon /is/ realistic) without proving 
IRIW at the same time? I wonder how that's possible. Or what gets 
forsaken. Transitive closure goes? Ok, I see that happening; but then it 
is not about total order as such.

Alex

>
> ---
> Marko

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