Re: user-defined op= for type with reference member

From:

James Kanze <james.kanze@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Mon, 21 Jun 2010 15:46:39 -0700 (PDT)

Message-ID:

<a28d1a29-e66b-4937-be0a-ff79dd639bdd@y4g2000yqy.googlegroups.com>

On Jun 11, 10:27 pm, "Leigh Johnston" <le...@i42.co.uk> wrote:

"Paul Bibbings" <paul.bibbi...@gmail.com> wrote in message

news:87iq5pe056.fsf@gmail.com...

Since I have used the following in another post, can someone
just confirm (or otherwise) whether the following definition
of a user-defined op= for a type with a reference member is
well defined?

  class AType
  {
  public:
     AType(int& i)
        : i_(i)
     { }
     // ...
     AType& operator=(const AType& other)
     {
        if (this != &other)
        {
           this->~Atype();
           new (this) AType(other);
        }
        return *this;
     }
  private:
     int& i_;
  };

According to my reading of the example given in [basic.life]
=A73.8/7 I believe that it is, in this instance (since the
constructor doesn't throw, except on bad_alloc).

Attempting to use this trick to reseat a reference or
destruct/re-construct const members is UB IIRC, there was a
thread about this a few months ago I think.

It's not undefined behavior per se. The standard is quite
clear that it is defined, and what its semantics are.

The problem is that its defined semantics can lead to undefined
behavior in so many cases that it's just not worth the bother;
the risk is too great. (And the undefined behavior can appear
as a result of some future evolution in the code.)

With regards to the orginal poster's problem: the answer is
simple: if you want to support assignment, use a pointer, rather
than a reference.

--
James Kanze