#### 12.7.4 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;
           A():Base("A"){}
         };

         struct B:virtual Base{
           int bval;
           B():Base("B"){}
         };

         struct Derived:public A, public B{
           Derived():Base("Derived"){}
         };

         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 more than once. 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.