[Bug libstdc++/15910] can't compile self defined void distance(std::vector<T>, std::vector<T>)

["adah at netstd dot com" <gcc-bugzilla at gcc dot gnu dot org>]

------- Additional Comments From adah at netstd dot com  2005-08-12 10:50 -------
(In reply to comment #97)
> Subject: Re:  can't compile self defined void distance(std::vector<T>, 
std::vector<T>)
> "adah at netstd dot com" <gcc-bugzilla@gcc.gnu.org> writes:
> | > you still fail to provide such a definition for "interface".
> | > -- Gaby
> | 
> | Why should I?
> Because, it is *you* who brought that topic in this discussion.
> |  Herb defined it in Exceptional C++, Item 32:
> | 
> | For a class X, all functions, including free functions, that both
> | 
> | * `mention' X
> | * are `supplied with' X
> | 
> | are logically part of X, because they form part of the interface of X.
> That definition bans std::swap as currently usable
> with std::complex<double>, e.g. 
>    std::complex<double> w, z;
>    swap(z, w);

Herb's opinion is that one should write `std::swap(...)'.  `Swap' is just an 
algorithm, and has nothing to do with `complex' per se.  It is clearly stated 
in Section 4.3 of his proposal.

> You would have hard-time convincing me that I want that behaviour for
> standard C++, let alone GCC.  Furtheremo more, you have to define
> "supplied with".  If N gives me
>      namespace N {
>         struct X { };
>      }
> and M adds
>     namespace M {
>        using N::X;
>        f(X);
>     }
> is M::f part of N::X's interface?

I guess `No'.  Herb's rationale is in C++ Coding Standards:

57. Keep a type and its nonmember function interface in the same namespace.

Please differentiate between what I request and what we discuss.  I did not 
advocate to implement Herb's proposal here.  At least not yet.

Best regards,

Yongwei


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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=15910