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

From:
"Leigh Johnston" <leigh@i42.co.uk>
Newsgroups:
comp.lang.c++
Date:
Fri, 11 Jun 2010 22:43:41 +0100
Message-ID:
<kcadnVLhL5mXM4_RnZ2dnUVZ8uqdnZ2d@giganews.com>
"Paul Bibbings" <paul.bibbings@gmail.com> wrote in message
news:87631pdz6o.fsf@gmail.com...

"Leigh Johnston" <leigh@i42.co.uk> writes:

"Pete Becker" <pete@versatilecoding.com> wrote in message
news:2010061111295323604-pete@versatilecodingcom...

On 2010-06-11 11:21:09 -1000, Paul Bibbings said:

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).


Yes, it's well-defined, but it's a really bad idea:

class BType : public AType
{
public:
BType& operator=(const BType& other)
{
return AType::operator==(other);
}
};

BType b1, b2;
b2 = b1; // nasty

In this particular example, there's almost certainly no nasty
behavior. But add a virtual function to AType (and make AType's
destructor virtual) and override the function in BType.

AType *at = &b2;
at->virtual_function();

Now things are messy, because the code says that b2 has type BType,
but BType's constructor has not been called.

--
 Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)


It is well defined in the sense that it is defined to be undefined
behaviour. You cannot reseat references.


But you are not reseating a reference, surely. The above code -
notwithstanding Pete's comments which I am still digesting - destructs a
constructed object and creates a new one in its place. That is a very
different beast, wouldn't you say?

Regards

Paul Bibbings


It is UB, see my reply else-thread.

/Leigh

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