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

Pete Becker <>
Fri, 11 Jun 2010 11:29:53 -1000
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
      AType(int& i)
         : i_(i)
      { }
      // ...
      AType& operator=(const AType& other)
         if (this != &other)
            new (this) AType(other);
         return *this;
      int& i_;

According to my reading of the example given in [] ?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
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;

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

Roundhouse Consulting, Ltd. ( Author of "The
Standard C++ Library Extensions: a Tutorial and Reference

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