> On Mar 26, 2004, at 7:45 PM, David Holmes wrote:
> > "roach motel" rules are exact/complete. ie if my code contains no
> > thread-local volatiles or thread-local sync then I CAN explain the
> > behaviour in terms of the "barriers" (memory AND code-motion) inherent
> > "roach motel" rules. Right?
> Initially, x = y = v = 0, v is volatile
> Thread 1:
> r3 = y
> v = 1
> r2 = v
> x = 1
> Thread 2:
> r1 = v
> v = r1+1
> Thread 3:
> r4 = x
> y = r4
> Execution in question:
> r1 = 1, r2 = 2, r3 == r4 == 1
> OK, so v is not thread local, and so it can't be removed.
> Roach motel order would prevent reordering the read of y or the write
> of x in thread 1.
That's fine. r3 is not guaranteed to read a particular value of y. The write
to x is not guaranteed to be visible to any other thread - but could be.
> But the ordering constraints on x and y are only visible to thread 2,
> not thread 3.
Yes. There is no synchronization in thread 3 with any other thread - so it
can see anything for x. This is not a correctly synchronized program.
> So it is possible for thread 3 to see the write to x by thread 1, and
> for thread 1 to see the write to y by thread 3.
Now you lost me again. If thread 3 sees the write to x by thread 1 then:
t1: x = 1 hb t3: r4 = x
And if thread 1 sees the write to y by thread 3 then:
t3: y = r4 hb t1: r3 = y
But t3: r4 = x hb t3: y = r4 because there is a data dependency, so
program order has to be obeyed.
And t1: r3 = y hb t1: x = 1 because of the volatile read and write of v
And that's a contradiction.
Is that you point? That the statements in thread 1 can be reordered? If so
what are the allowed reorderings?
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