question on polymorphism

This is a discussion on question on polymorphism within the C++ Programming forums, part of the General Programming Boards category; ok..can anyone tell me more in detail what polymorphism is..all i know so far is that the base class and ...

  1. #1
    samurai warrior nextus's Avatar
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    Question question on polymorphism

    ok..can anyone tell me more in detail what polymorphism is..all i know so far is that the base class and derive class has the same member functions name declare with virtual it gives it dynamic linkage..and kind of same with pointers....whats so good about ..please give examples..thanks
    nextus, the samurai warrior

  2. #2
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    Ok... I will try :)

    So imagine that you have this class

    Code:
    class Base
    {
        int x;
        
    public:
        Base (int xx=0): x(xx){}
        virtual int getNum() {return x;};
        void PrintNum( cout<<x;);
    };
    
    class Derivated
    {
        int newX;
    public:
        Derivated (int xx=1, int y=5): Base(xx){newX=y;}
    
        int getNum() {return newX;}
        void PrintNum() {cout<<NewX;}
    };
    
    void main()
    {
        Base num(5);
        Derivated num2(5,10);
        Base *ptr;
    
        int y=num.getNum();   //just call the getNum from the base
        cout<<y;  //y=5
        num.PrintNum();  //print 5
    
        int w=num.getNum();  //Now it will call the getNum from the derivated classe because this function is virtual
        cout<<w;  //w=10
      
        //Now came the nice part
        ptr=new Derivated(5,20);
    
        int z=ptr.getNum(); //it will call the getNum from the derivated
        cout<<z;  //z=20
        ptr.printNum();  //it will print 5
    }
    i hope that i didn't say any stupid thing . And i hope that you understand a bit more from classes...

  3. #3
    samurai warrior nextus's Avatar
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    your Derivated class didnt inherit any data members from the Base class

    Code:
    class Derivated
    
    // should be
    
    class Derivated: public Base
    nextus, the samurai warrior

  4. #4
    samurai warrior nextus's Avatar
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    i fix your code

    Code:
    #include <iostream>
    
    using namespace std;
    
    class Base
    {
        int x;
        
    public:
        Base (int xx=0): x(xx){}
        virtual int getNum() {return x;};
        void PrintNum() { cout<<x; }
    };
    
    class Derivated: public Base
    {
        int newX;
    public:
        Derivated (int xx=1, int y=5): Base(xx){newX=y;}
    
        int getNum() {return newX;}
        void PrintNum() {cout<<newX;}
    };
    
    void main()
    {
        Base num(5);
        Derivated num2(5,10);
        Base *ptr;
    
        int y=num.getNum();   //just call the getNum from the base
        cout<<y;  //y=5
        num.PrintNum();  //print 5
    
        int w=num.getNum();  //Now it will call the getNum from the derivated classe because this function is virtual
        cout<<w;  //w=10
      
        //Now came the nice part
        ptr=new Derivated(5,20);
    
        int z=ptr->getNum(); //it will call the getNum from the derivated
        cout<<z;  //z=20
        ptr->PrintNum();  //it will print 5
    }
    now it is syntax correct
    nextus, the samurai warrior

  5. #5
    samurai warrior nextus's Avatar
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    also it should be num2.getNum() // outputs 10. but thanks i get what it does now.

    if the derive class has the same member function name as the base class it will have static linkage which it will call the base class member function, but if you use virtual then it has dynamic linkage which then when you call through a pointer it will chose the object that is associated with it.
    nextus, the samurai warrior

  6. #6
    Registered User
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    Talking Of course

    Sure it must be class Derivaterublic Bse. Sorry...

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