# Re: [ub] Justification for < not being a total order on pointers?

From: Gabriel Dos Reis <gdr_at_[hidden]>
Date: Thu, 17 Oct 2013 18:16:02 -0500
Lawrence Crowl <Lawrence_at_[hidden]> writes:

| On 10/17/13, James Dennett <jdennett_at_[hidden]> wrote:
| > On Oct 17, 2013, Gabriel Dos Reis <gdr_at_[hidden]> wrote:
| > > James Dennett <jdennett_at_[hidden]> writes:
| > > > I think it's not useful though; nothing guarantees that
| > > > p == q => intptr_t(p) == intptr_t(q), so far as I know,
| > >
| > > How do you arrive to this conclusion?
| >
| > As always, it's hard to point to where something is _not_
| > specified.
| >
| > I'm aware of no such guarantee. If you're aware of one, you
| > can presumably cite it, and I'd appreciate such a citation.
| > All I see is "A pointer can be explicitly converted to any
| > integral type large enough to hold it. The mapping function is
| > implementation-defined." We can debate whether the word
| > "function" there should be taken in a pure mathematical sense,
| > and whether "a pointer" means the value rather than the
| > representation, and if we have just the right reading (it's a
| > function on pointer values) then we get that equal pointers
| > have equal values when converted to intptr_t. We don't have
| > any guarantee about whether it's an order-preserving function
| > though, unless that's somewhere else.
|
| As an example, consider 8086 segmented pointers A:B and C:D that
| point to the same object. If the intptr_t conversion consistes
| of concatentationg A and B into a 32-bit value, the implication
| above is broken. To preserve the implication, the implementation
| would need to split and shift A and then do some carry adds,
| which is much more expensive than the simple concatenation.

Let's postulate that segmented architecture do not exist anymore; where
are we in terms of the implication?

-- Gaby