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Subject: Re: [ub] Justification for < not being a total order on pointers?
From: Nevin Liber (nevin_at_[hidden])
Date: 2013-08-26 11:12:15


On 26 August 2013 11:00, Jeffrey Yasskin <jyasskin_at_[hidden]> wrote:

>
> Could someone explain why we need to allow operator<(T*) to be a non-order?
>

It comes from C. I believe it comes from the days of segmented
architectures.

I do not know of any modern machines that have such architectures and have
C++11 compilers for them. Whenever it comes up for discussion on various
reflectors, no one has mentioned one either. I for one would like to see
this restriction go away.

Armchair thought: maybe we should propose a total ordering for pointers
(for C++17 at this point) and see if anyone objects?

All that being said, I believe Library is inconsistent in its use of
operator< vs. std::less<T>, and that needs to be addressed separately.
Pointers are the current poster child for the issue but user code might be
specializing std::less as well.

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



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