I have a question on C++ double dispatch. In the code below, I want the results from the second set to match the results from the first set.
I don\'t know the actua
Thank you so much. This solved my problem and I understand what double dispatch means! Here is the complete code for posterity (somebody please teach me how to get the formatting right):
#include
using namespace std;
class BaseClass;
class Derived1;
class Derived2;
class Derived3;
class Processor {
public:
Processor(){}
virtual void processObj(BaseClass* bc);
virtual void processObj(Derived1* d1);
virtual void processObj(Derived2* d2);
};
class BaseClass{
public:
BaseClass(){}
virtual void ProcessThis(Processor &p) { p.processObj(this); }
virtual void myFunction(){cout << "base myFunction called" << endl;}
};
class Derived1: public BaseClass{
public:
Derived1():BaseClass(){}
void ProcessThis(Processor &p) { p.processObj(this); }
void myFunction(){cout << "Derived1 myFunction called" << endl;}
};
class Derived2: public BaseClass{
public:
Derived2():BaseClass(){}
void ProcessThis(Processor &p) { p.processObj(this); }
void myFunction(){cout << "Derived2 myFunction called" << endl;}
};
class Derived3: public BaseClass{
public:
Derived3():BaseClass(){}
void ProcessThis(Processor &p) { p.processObj(this); }
void myFunction(){cout << "Derived3 myFunction called" << endl;}
};
void Processor::processObj(BaseClass* bc){cout << "got a base object" << endl; bc->myFunction();}
void Processor::processObj(Derived1* d1){cout << "got a derived1 object" << endl; d1->myFunction();}
void Processor::processObj(Derived2* d2){cout << "got a derived2 object" << endl; d2->myFunction(); }
int main() {
BaseClass *bcp=new BaseClass();
Derived1 *dc1p=new Derived1();
Derived2 *dc2p=new Derived2();
Derived3 *dc3p=new Derived3();
Processor p;//can also use Processor* p = new Processor()
//first set results
bcp->ProcessThis(p);
dc1p->ProcessThis(p);
dc2p->ProcessThis(p);
dc3p->ProcessThis(p);
BaseClass *bcp1=bcp;
BaseClass *dc1p1=dc1p;
BaseClass *dc2p1=dc2p;
BaseClass *dc3p1=dc3p;
//second set results
bcp1->ProcessThis(p);
dc1p1->ProcessThis(p);
dc2p1->ProcessThis(p);
dc3p1->ProcessThis(p);
Processor p2;
bcp1->ProcessThis(p2);
dc1p1->ProcessThis(p2);
dc2p1->ProcessThis(p2);
dc3p1->ProcessThis(p2);
return 0;
}