How to make this operator= overload work? - c++

Question

I was experimenting with making the operator= be virtual, like so:
#include <iostream>
#include <string>
class Base {
public:
std::string field1;
Base(std::string str) : field1(str) {}
virtual Base& operator=(const Base& other)
{
field1 = other.field1;
return *this;
}
};
class Derived : public Base {
public:
std::string field2;
Derived(std::string str1, std::string str2) : Base(str1), field2(str2) {}
virtual Derived& operator=(const Derived& other) override
{
Base::operator=(other);
field2 = other.field2;
return *this;
}
};
However, this gives a compiler error, because the Derived function is not actually overloading anything, the signatures are different.
Is it possible to override the operator= to write code like this?
Base* ptr = new Derived("old 1", "old 2");
Derived derived("new 1", "new 2");
*ptr = derived; // <- use the derived class operator= to assign both field1 and field2

Answer 1

This operator
virtual Derived& operator=(const Derived& other);
does not override the copy assignment operator declared in the base class.
You have to write
Derived& operator=(const Base& other) override;
That is the type of the parameter shall be const Base &.
Here is a demonstrative program.
#include <iostream>
struct A
{
virtual A & operator =( const A & )
{
std::cout << "A::operator =\n";
return *this;
}
};
struct B : A
{
virtual B & operator =( const A &a ) override
{
A::operator =( a );
std::cout << "B::operator =\n";
return *this;
}
};
int main()
{
B b1;
A &rb1 = b1;
B b2;
b2 = rb1;
return 0;
}
Its output is
A::operator =
B::operator =