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

From: Gabriel Dos Reis <gdr_at_[hidden]>
Date: Thu, 17 Oct 2013 09:28:12 -0500
Nevin Liber <nevin_at_[hidden]> writes:

| On 16 October 2013 10:06, Gabriel s too subtle for even most committee members,
| let alone mere mortals, to get right.Dos Reis <gdr_at_[hidden]> wrote:
|
|
| | I would not have trouble telling people (especially notices):
| "Ignore expect
| | talks
| | about operator< on pointers. Prefer std::less<T>, unless you meant
| a
| | relationship
| | between objects pointed to, in which use operator<. Mean what you
| say
| | and say
| | what you mean."
| |
| |
| |
| | So what about std::less<void>? Should people be using it?
|
| What is wrong about it?
|
|
| It requires operator<. If people specialize std::less<T> instead adding an
| operator<, it won't work. It would be surprising that:
|
| set<T, less<T>> works, but
| set<T, less<>> does not.
|
| One of the two motivations for this in n3421 is to not require that the
| argument type in a comparator functor be explicitly specified, but that assumes
| there is an operator< for the type, or possible compiler magic for pointer
| comparisons because of the ub issue.

If your goal is to make std::less an alternate syntax for operator<,
then from that logic, it would not make much difference to require
std::less<void> to use std::less<T> instead of operator<. Therefore,
there would be no difference between

   set<T, less<T>>
  | set<T, less<>>


-- Gaby

Received on 2013-10-17 16:28:29