重载基类及其派生类的输出运算符（C ++）

Question

this is a question on overloading the output operator in C++: How do you overload << on BOTH the base class and the derived class? Example:

#include <iostream>
#include <cstdlib>
using namespace std;


class Base{
public:
virtual char* name(){ return "All your base belong to us.";}
};

class Child : public Base{
public:
virtual char* name(){ return "All your child belong to us.";}
};


ostream& operator << (ostream& output, Base &b){
output << "To Base: " << b.name() ;
return output;}

ostream& operator << (ostream& output, Child &b){
output << "To Child: " << b.name() ;
return output;}


int main(){

Base* B;
B = new Child;

cout << *B << endl;

return 0;

}

The output is

To Base: All your child belong to us."

so the name() in Child shadowed that of Base; but the overloading of << does not descend from Base to Child. How can I overload << such that, when its argument is in Base it uses ONLY the Base version of << ?

I want "To Child: All your child belong to us." to be output in this case.

Answer 1

Make it a virtual function.

class Base{
public:
  virtual const char* name() const { return "All your base belong to us.";}
  inline friend ostream& operator<<(ostream& output, const Base &b)
  {
    return b.out(output);
  }
private:
  virtual ostream& out(ostream& o) const { return o << "To Base: " << name(); }
};

class Child : public Base{
public:
  virtual const char* name() const { return "All your child belong to us.";}
private:
  virtual ostream& out(ostream& o) const { return o << "To Child: " << name(); }
};

Answer 2

int main(){

Base* B;
B = new Child;

cout << ( Child &) *B << endl;

return 0;

}