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9.9.3 Implicit Copy-Assignment for Virtual Bases

When a base class is virtual, only one subobject of the base class belongs to each full object. Also, the constructors and destructors are invoked only once, and called from the most-derived class. However, such objects behave unspecified when being assigned. For example:

struct Base{
  char *name;
  Base(char *n) : name(strdup(n)){}
  Base& operator= (const Base& other){
   free (name);
   name = strdup (other.name);

struct A:virtual Base{
  int val;

struct B:virtual Base{
  int bval;

struct Derived:public A, public B{

void func(Derived &d1, Derived &d2)
  d1 = d2;

The C++ standard specifies that `Base::Base' is only called once when constructing or copy-constructing a Derived object. It is unspecified whether `Base::operator=' is called more than once when the implicit copy-assignment for Derived objects is invoked (as it is inside `func' in the example).

g++ implements the "intuitive" algorithm for copy-assignment: assign all direct bases, then assign all members. In that algorithm, the virtual base subobject can be encountered many times. In the example, copying proceeds in the following order: `val', `name' (via strdup), `bval', and `name' again.

If application code relies on copy-assignment, a user-defined copy-assignment operator removes any uncertainties. With such an operator, the application can define whether and how the virtual base subobject is assigned.

This document was generated by Charlie & on June, 17 2001 using texi2html