Subject: Re: [ub] Justification for < not being a total order on pointers?
From: James Dennett (jdennett_at_[hidden])
Date: 2013-10-17 15:07:15
On Thu, Oct 17, 2013 at 12:41 PM, Gabriel Dos Reis <gdr_at_[hidden]> wrote:
> James Dennett <jdennett_at_[hidden]> writes:
> | On Thu, Oct 17, 2013 at 8:19 AM, Gabriel Dos Reis <gdr_at_[hidden]> wrote:
> | > Nevin Liber <nevin_at_[hidden]> writes:
> | >
> | > | But, even if segmented architectures, unlikely though it is, do come back, the
> | > | ordering problem still has to be addressed, as std::less<T*> is required to
> | > | totally order pointers. I just want operator< to be an alternate spelling for
> | > | that property.
> | >
> | > I will repeat this again: std::less<T*> is absolutely not a problem,
> | > because this
> | >
> | > return intptr_t(p) < intptr_t(q);
> | >
> | > is valid and portable definition that requires no other special handling.
> | 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.
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