Tim Daly <[email protected]> writes: | Note that, under this definition | | a:Dequeue(Integer):= dequeue [1,2,3,4,5] | b:Queue(Integer):= queue [1,2,3,4,5] | | hash(a) == hash(b) ^^ I suspect you meant '='.
In any case, unless you have a hash function that is a bijection of the whole universe onto the finite set SingleInteger, you're going to get different objects/values mapped to the same hash value no matter what. OpenAxiom being strongly typed, I'm not worried that you're going to put both a and b in the same HashTable and not being able to avoid the collision. -- Gaby _______________________________________________ Axiom-developer mailing list [email protected] http://lists.nongnu.org/mailman/listinfo/axiom-developer
