2 Oct
2011
2 Oct
'11
9:38 p.m.
Also, being a singular iterator is independent of type, while being a random-access iterator is determined by type.
No again. Objects of this type are not singular iterators:
struct nonsingular { private: void operator=(nonsingular const&); };
Singular values crop up in all kinds of contexts, BTW. Do ints support division? Well, yes, unless the denominator is zero.
I'm sorry. This is completely wrong. 0 is value -- a state of an numeric type -- just like singularity is the state of some iterators. In fact, I think you actually prove Chris' point, here; you've just extended the notion of singularity from iterators to integers. I would take that to mean that singularity is independent of type.