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#### 13.7.4 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.