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.
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Licensed under the GNU Free Documentation License, Version 1.3.
https://gcc.gnu.org/onlinedocs/gcc-9.3.0/gcc/Copy-Assignment.html