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virtual ineritance

Mor1994 (37)
Hi everyone!
i have a question- when i write: Base* b=new Multiple(); b->print();
why does Base::print() activated, and not Multuple::print()?
print is virtual function..
thanks for help !

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  class Base
    {
        public:
       virtual void print(){}
    };
    class DerivedOne:  virtual public Base
    {
        public:     
        void print() const {cout << "DerivedOne\n";}
    };
    class DerivedTwo :virtual public Base
    {
        public:
        void print() const {cout << "DerivedTwo\n";}     
    }; 
    class Multiple : public DerivedOne, public DerivedTwo
    {
        public:
        void print() const {DerivedTwo :: print();} 
		void g(){cout<<"blabla";}
    };
Last edited on
Yanson (885)
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  class Base
    {
        public:
       virtual void print() { }
    };



The base classes print method needs to be const

virtual void print() const { }


Also you can make use of the override keyword which gives an error when your method fails to override a base class method.

Example
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#include <iostream>

using namespace std;
  class Base
    {
        public:
       virtual void print() {}
    };
    class DerivedOne:  virtual public Base
    {
        public:
        void print() const {cout << "DerivedOne\n";}
    };
    class DerivedTwo :virtual public Base
    {
        public:
        void print() const {cout << "DerivedTwo\n";}
    };
    class Multiple : public DerivedOne, public DerivedTwo
    {
        public:
        void print() const override {DerivedTwo::print();}
		void g(){cout<<"blabla";}
    };
Mor1994 (37)
thanks, but this code was written like this on purpose.
can you explain again why Base::print is called? even thogh print is virtual..?
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