Subject: Re: [ub] Justification for < not being a total order on pointers?
From: Ville Voutilainen (ville.voutilainen_at_[hidden])
Date: 2013-10-16 10:50:57
On 16 October 2013 18:46, Gabriel Dos Reis <gdr_at_[hidden]> wrote:
> | My own personal view (not that of chair) is that if
> std::less<T>(l,r) and
> | "l < r" are
> | both defined, then they should yield the same answer.
> | Which fails for pointers.
> "fails" in which sense? It is certainly true in the current standards
> these expressions are both defined when 'l' and 'r' are related
> addresses (relative to the same object), which was exactly my point.
Fails in the sense that less<int*>(l, r) and l<r do not necessarily yield
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