Questiopn on Pure Virtual Function

Can I implement a pure virtual function?
How?
TSENTHILKLUMARAsked:
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abesoftConnect With a Mentor Commented:
You can indeed implement a pure virtual function.  A PV function is just like any other virtual function, with one exception: no class containing (or inheriting without overriding) a pure virtual function cannot be instantiated.

So, for example:
class PVBase{
public:
    virtual void Print() = 0
    {   cout << "Base printing support";}
};
class Derived: public PVBase{
    virtual void Print()
    {   cout << "Derived printing support";
        // Now, print the base portion!
        PVBase::Print();
    }
};

If you choose to implement your pure virtual function, then it can still be called, although it won't be called "virtually".  That is, you will never end up calling it by:
    myPointer->Print()
because myPointer will (by definition) be an instance of a class that has over-ridden the Print function.

I think that this is a rather obscure part of C++, but it can be used effectively in some cases.  Basically, this mechanism allows the base class to provide some "standard" operations for a function, but still require derived classes to specialize the behaviour.  Printing is one place that I could imagine it being useful.

To alcindor: Yes, you are right in saying that you will never _need_ to implement a PV.  But it is nevertheless useful, in that it can be called.
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daknight2000Commented:
yes u can
somehting like
virtual myfunction()=0
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KangaRooCommented:
class A
{
     virtual void f();  // regular virtual
     virtual void v() = 0; // pure virtual
};

void A::f()
{
    cout << "A::f()" << endl;
}

void A::v()
{
    cout << "A::v() will never be executed" << endl;
}
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TSENTHILKLUMARAuthor Commented:
what i would like to know is in multilevel inheritance if i call a pure virtualfunction from a derivedclass function whichis not virtual it is getting executed which should not be, as a pure virtual function can not have a body
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Jan LouwerensSoftware EngineerCommented:
once a function is declared as virtual, it is virtual for every class that derives from that base class, whether or not the keywork virtual is actually used in the derived class. And since you can't instantiate an object of a class that has any pure virtual functions, you can never run the risk of calling a virtual function that contains no body
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WynCommented:
How can you call a pure vitual function ?
Base:Vitual()??
It's not legal and you'd never do that...
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kishore_joshiCommented:
class Base
{
      int i;
public:
      virtual void f() = 0;      // Pure Virtual Function
};


void Base::f()             // Implimentation of Pure Virt Function
{
      cout<<" \n I cannot be called.... ";
}


class Derv1:public Base
{
      int j;
public:
      void f()
      {
            cout<< " I am in Derv1 \n ";
      }

      void f_d1()
      {
       cout<<" in f_d1 function of the d1 class.. going to call base PVF";
       Base::f();   // Calling the pure virtual fuction of the Base class
      }
};


class Derv2:public Derv1
{
      int k;
public:
      void f()
      {
            cout<< " I am in Derv2 \n ";
      }

};



void main()
{

      Base *bPtr;

      Derv1 d1; bPtr = &d1;
      bPtr->f();
      d1.f_d1();


      Derv2 d2; bPtr = &d2;
      bPtr->f();
      d2.f_d1();

}

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alcindorCommented:
If the original question still stands, then I would say that you wil NEVER need to implement a pure virtual function. If you wish to provide some default base function then don't make it pure virtual.
The idea of pure virtual functions are to define interface specifications for derived classes.

Roger
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abesoftCommented:
Oh, the other part of your question was "how do I implement a PV function"?  The syntax is the same as any virtual function.  You can define it inline (as in my example) or out-of-line, as in kishore_joshi's example.  The only difference is the "=0" item in the class header.
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