Virtual Function Question

the derived class doesnt need the virtual keyword

pure meens the the function doesnt exsist and needs to be created in a derived class

classes with pure virtual functions cant be created

If you declare a function virtual in a base class, overrides of that function implicitly declared virtual in all derived classes. (However, it's bad form to leave off the virtual in the derived classes.)

A pure virtual function must be overridden in derived classes in order for the derived classes to be instantiated. A class with any non-overriden pure virtual functions is called an abstract class.

Regarding your first question: there is no difference. The method will be implicitly virtual, whether you add the keyword or not.

WRT your second question: Yes, the '= 0' makes it a pure virtual function and the difference is that a pure virtual function makes the class abstract, that is it cannot be instanciated. Any derived classes that do not provide an implementation for all pure virtual methods of the base class will also be abstract classes. You can use this behaviour to force the use of derived classes.

Your third question: upgrade to VC.NET! NOW! Just kidding [smile]

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Too slow... [sad]
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Quote:

Original post by mike25025 the derived class doesnt need the virtual keyword

Unless you plan to have classes derived from it able to override the DoSomething function as well. Aka:

struct A //struct A == class A { public:{    virtual void DoSomething() = 0; //MUST be implemented in derived class. This class CAN NOT be created.};struct B : public A{    virtual void DoSomething(); //MAY be implemented in derived class. This class CAN be created.};struct C1 : public B{    void DoSomething(); //SHOULD NOT be implemented in derived class, most likely... reason is explained bellow. CAN be created.};struct C2 : public A{    void DoSomething(); //Same as C1};struct D : public C1{    void DoSomething(); //This is were things get interesting...};//Now, the difference, illustrated in a function:void function( void ){    D my_d;    my_d.DoSomething(); //Calls D::DoSomething() as expected.    A * my_ptr = new D;    my_ptr->DoSomething(); //Calls C1::DoSomething(), probably not expected.    delete my_ptr;    my_ptr = new B;    my_ptr->DoSomething(); //Calls B::DoSomething() as expected.    delete my_ptr;}

Quote:

pure meens the the function doesnt exsist and needs to be created in a derived class

It is complete valid in C++ to do the following:

class CBaseClass{public:    virtual void DoSomething() = 0;};void CBaseClass::DoSomething(){    //...}

The class will not be createable and derived classes will need to implement DoSomething(). However, this allows the following code:

class CDerviedClass : public CBaseClass{public:    virtual void DoSomething();}void CDerivedClass::DoSomething( void ){    CBaseClass::DoSomething(); //*    //do more stuff...}

Quote:

classes with pure virtual functions cant be created

edit: unbrokened forum.

* We're "adding onto", not "replacing" DoSomething() in this case. In reality we're replacing DoSomething, but we're calling the base class's DoSomething() first, so it has the same effect.

Quote:

Original post by MaulingMonkey Quote: Original post by mike25025 the derived class doesnt need the virtual keyword Unless you plan to have classes derived from it able to override the DoSomething function as well.

No, once virtual, always virtual. You will get exactly the same effect if you write virtual in the child or intermediate base as long as the original declaration was virtual. It's entirely optional to put virtual in a child once the function is declared virtual in a base, and it never changes the functionality of the code if you do or do not continue writing virtual in each override.