on Sun Oct 02 2011, Andrew Sutton
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.
Yes. I don't see that as a contradiction.
In fact, I think you actually prove Chris' point, here; you've just extended the notion of singularity from iterators to integers.
It's not like I just did something new, here!
I would take that to mean that singularity is independent of type.
Although many types have singular values with respect to certain operations (NULL is singular with respect to pointer dereference), - Not every type has singular values - Some types have multiple singular values - In general, one type's singular values are distinct from another type's singular values I don't know what you and Chris mean by "independent of type," but to me this sounds like singularity is highly dependent on type. I suspect this situation is less one of "you're completely wrong" than "you and I are understanding the same words in different ways." Cheers, -- Dave Abrahams BoostPro Computing http://www.boostpro.com