Hello,
En 19 Nov 2003 21:32:24 +0100, Ben Moseley <ben_moseley@mac.com> escribio:
> But isn't there observable nondeterminism in the declarative concurrent
> subset of Oz by exactly the same argument:
>
> declare C
> thread
> C=1
> end
> thread
> C=2
> end
>
> ... with this example you cannot tell in advance which value C will be
> bound to. Certainly the other binding attempt will fail, but isn't this
> observable nondeterminism nonetheless?
When you run this, the overall result is always failure because of
one-assignment variables, I guess always-failure could count as a
deterministic output ...
But, that is cheating, the problem is that the unification procedure
raises an exception, and the declarative model + exceptions is no longer
declarative as the book states.
-- Andres Becerra Sandoval - Please send submissions to users@mozart-oz.org and administriva mail to users-request@mozart-oz.org. The Mozart Oz web site is at http://www.mozart-oz.org/. Please send bug reports to bugs@mozart-oz.org.