At 03:06 PM 19/04/2004 -0400, Bill Pugh wrote:
>class A {
> static A g;
> long x;
> protected void finalize() {
> g = this;
> long r1 = x;
> // ...
> }
> }
>
>
> Thread 1:
> A r2 = new A();
> r2.x = 1;
> r2.x = -1;
>
> Thread 2:
> do {
> r3 = A.g;
> } while (r3 == null);
> r4 = r3.x;
>
>Are we guaranteed that r1 and r4 will both see the value -1?
>
>At a intuitive level, before running the finalizer on an object o,
>we need to make sure that if two conflicting writes w1 and w2
>have been performed to a field/element of o, and w1 happens-before
>w2, then we need to make sure that before the finalizer is run, the
>write w1 becomes no longer accessible.
Yet there is data race on the accesses to g, and no happens-before edge
that prevents r4 = 1. This outcome looks no different from other
counter-intuitive outcomes that arise from naive multi-threaded
programming. Is there any reason to treat this one as special?
As so often here, I'm left wondering whether I've missed the point.
Sylvia.
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