On 17 October 2013 23:13, Gabriel Dos Reis <gdr_at_[hidden]> wrote:

> | I suppose the other implication that is on the mind of many people
> but
>
| not being discussed is
> |
> | p != q => intptr_t(p) != intptr_t(q)
> |
> |
> | I'm not concerned about that, as p -> intptr_t(p) -> p has to result in a
> | matching pointer. I don't see how that is workable if the implication
> above
> | fails.
>
> Hmm, remember I am very slow. Please explain this for me in further
> details; I -think- I guess what you are saying but I would rather
> make sure you spell out for me how this is working on today's machines
> for which we need to make operator< a total ordering, and how they match
> (or differ from existing implementations.)
>

Proof by contradiction:

1. Suppose p != q => intptr_t(p) == intptr_t(q)

The guarantee for a pointer p of type T* is, if intptr_t is provided:

p equals (T*)(intptr_t)(p);

let x = intptr_t(p)

x must also equal (intptr_t)(q) because (intptr_t)(p) == (intptr_t)(q)

It is impossible for (T*)(x) != (T*)(x).

With substitution, (T*)(intptr_t)(p) != (T*)(intptr_t)(q) is also
impossible.

Therefore,

p != q => intptr_t(p) != intptr_t(q)

must be true.

Q.E.D.

-- 
 Nevin ":-)" Liber  <mailto:nevin_at_[hidden]>  (847) 691-1404