Subject: Re: [ub] Justification for < not being a total order on pointers?
From: Nevin Liber (nevin_at_[hidden])
Date: 2013-10-17 23:30:39
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
| 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
> | 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
p != q => intptr_t(p) != intptr_t(q)
must be true.
-- Nevin ":-)" Liber <mailto:nevin_at_[hidden]> (847) 691-1404
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