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

From:

Kai-Uwe Bux <jkherciueh@gmx.net>

Newsgroups:

comp.lang.c++

Date:

Fri, 11 Jun 2010 23:57:36 +0200

Message-ID:

<huubgh$p4u$1@news.doubleSlash.org>

Paul Bibbings wrote:

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] ?3.8/7 I
believe that it is, in this instance (since the constructor doesn't
throw, except on bad_alloc).

It is undefined behavior according to [3.8/7, item 3], which requires:

  ...
  the type of the original object is not const-qualified, and, if a class
  type, does not contain any non-static data member whose type is const-
  qualified or a reference type,
  ...

Since AType contains a non-static reference member, the trick does not work
for AType. Note that in the example in the standard, there is no non-static
reference member.

I know: it's a bummer since classes with reference members are about the
only cases, where one might even consider this trickery (which is poor form
anyway).

Best

Kai-Uwe Bux