Re: Why is overloading operator. (member operator) forbidden?

From:
=?iso-8859-1?q?Erik_Wikstr=F6m?= <eriwik@student.chalmers.se>
Newsgroups:
comp.lang.c++
Date:
14 May 2007 01:50:50 -0700
Message-ID:
<1179132650.060861.279960@p77g2000hsh.googlegroups.com>
On 14 Maj, 09:33, "dasca...@gmail.com" <dasca...@gmail.com> wrote:

On 14 mei, 08:12, Erik Wikstr=F6m <eri...@student.chalmers.se> wrote:

There are two reasons against it that I know of, the first being that
the general consensus in the standard committee is that it might
introduce too much trouble* , since a user would never know what will
happen if he uses the .-operator, the second is the problem you
mentioned about how to implement the normal behaviour. I know of no
convincing reason for allowing it since it really would not enable
behaviour that can't be simulated in other ways.


I was thinking of a few things in particular, mainly allowing
transparent forwarding of interfaces. Given variadic templates and an
overloadable operator. you could define a function as such:

template <typename R, typename... Argtypes>
auto operator.(std::string funcname, Argtypes... args) { return
memfun(this, new gen_send<R, ArgTypes...>(funcname); }
template <typename R, typename... Argtypes>
class gen_send {
  private:
    string funcname;
  public:
    R &operator()(Argtypes... arguments) {
      // serialize and send
      // wait for return value, return value
    }

};

that would allow transparently forwarding of the function called,
irrelevant of which function it was. Marshalling the return type might
cause one or two small bits of trouble but I think there's a fairly
easy way around that.

* In fact a recent proposal (n2200) suggests that all operators,
except . (member selection), :: (namespace) and .* (memberpointer),
should be overloadable both as members and non member (that includes
new and delete).


Wouldn't that cause trouble with typeid and especially sizeof?


Ah, I missed those. I didn't read the whole of the proposal, but I
don't think those were included (nor ?:).

--
Erik Wikstr=F6m

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