MFC Programmer's SourceBook : Thinking in C++

virtual base classes

To solve the first problem, you must explicitly disambiguate the function vf( ) by writing a redefinition in the class mi.

The solution to the second problem is a language extension: The meaning of the virtual keyword is overloaded. If you inherit a base class as virtual, only one subobject of that class will ever appear as a base class. Virtual base classes are implemented by the compiler with pointer magic in a way suggesting the implementation of ordinary virtual functions.

Because only one subobject of a virtual base class will ever appear during multiple inheritance, there is no ambiguity during upcasting. Here’s an example:

//: C22:MultipleInheritance2.cpp
// Virtual base classes
#include <iostream>
#include <vector>
#include "../purge.h"
using namespace std;

class MBase {
public:
  virtual char* vf() const = 0;
  virtual ~MBase() {}
};

class D1 : virtual public MBase {
public:
  char* vf() const { return "D1"; }
};

class D2 : virtual public MBase {
public:
  char* vf() const { return "D2"; }
};

// MUST explicitly disambiguate vf():
class MI : public D1, public D2 {
public:
  char* vf() const { return D1::vf();}
};

int main() {
  vector<MBase*> b;
  b.push_back(new D1);
  b.push_back(new D2);
  b.push_back(new MI); // OK
  for(int i = 0; i < b.size(); i++)
    cout << b[i]->vf() << endl;
  purge(b);

} ///:~

The compiler now accepts the upcast, but notice that you must still explicitly disambiguate the function vf( ) in MI; otherwise the compiler wouldn’t know which version to use.

The "most derived" class and virtual base initialization

The use of virtual base classes isn’t quite as simple as that. The above example uses the (compiler-synthesized) default constructor. If the virtual base has a constructor, things become a bit strange. To understand this, you need a new term: most-derived class .

The most-derived class is the one you’re currently in, and is particularly important when you’re thinking about constructors. In the previous example, MBase is the most-derived class inside the MBase constructor. Inside the D1 constructor, D1 is the most-derived class, and inside the MI constructor, MI is the most-derived class.

When you are using a virtual base class, the most-derived constructor is responsible for initializing that virtual base class. That means any class, no matter how far away it is from the virtual base, is responsible for initializing it. Here’s an example:

//: C22:MultipleInheritance3.cpp
// Virtual base initialization
// Virtual base classes must always be
// Initialized by the "most-derived" class
#include <iostream>
#include <vector>
#include "../purge.h"
using namespace std;

class MBase {
public:
  MBase(int) {}
  virtual char* vf() const = 0;
  virtual ~MBase() {}
};

class D1 : virtual public MBase {
public:
  D1() : MBase(1) {}
  char* vf() const { return "D1"; }
};

class D2 : virtual public MBase {
public:
  D2() : MBase(2) {}
  char* vf() const { return "D2"; }
};

class MI : public D1, public D2 {
public:
  MI() : MBase(3) {}
  char* vf() const {
    return D1::vf(); // MUST disambiguate
  }
};

class X : public MI {
public:
  // You must ALWAYS init the virtual base:
  X() : MBase(4) {}
};

int main() {
  vector<MBase*> b;
  b.push_back(new D1);
  b.push_back(new D2);
  b.push_back(new MI); // OK
  b.push_back(new X);
  for(int i = 0; i < b.size(); i++)
    cout << b[i]->vf() << endl;
  purge(b);

} ///:~

As you would expect, both D1 and D2 must initialize MBase in their constructor. But so must MI and X, even though they are more than one layer away! That’s because each one in turn becomes the most-derived class. The compiler can’t know whether to use D1’s initialization of MBase or to use D2’s version. Thus you are always forced to do it in the most-derived class. Note that only the single selected virtual base constructor is called.

"Tying off" virtual bases with a default constructor

Forcing the most-derived class to initialize a virtual base that may be buried deep in the class hierarchy can seem like a tedious and confusing task to put upon the client programmer of your class. It’s better to make this invisible, which is done by creating a default constructor for the virtual base class, like this:

//: C22:MultipleInheritance4.cpp
// "Tying off" virtual bases
// so you don't have to worry about them
// in derived classes
#include <iostream>
#include <vector>
#include "../purge.h"
using namespace std;

class MBase {
public:
 // Default constructor removes responsibility:
  MBase(int = 0) {}
  virtual char* vf() const = 0;
  virtual ~MBase() {}
};

class D1 : virtual public MBase {
public:
  D1() : MBase(1) {}
  char* vf() const { return "D1"; }
};

class D2 : virtual public MBase {
public:
  D2() : MBase(2) {}
  char* vf() const { return "D2"; }
};

class MI : public D1, public D2 {
public:
  MI() {} // Calls default constructor for MBase
  char* vf() const {
    return D1::vf(); // MUST disambiguate
  }
};

class X : public MI {
public:
  X() {} // Calls default constructor for MBase
};

int main() {
  vector<MBase*> b;
  b.push_back(new D1);
  b.push_back(new D2);
  b.push_back(new MI); // OK
  b.push_back(new X);
  for(int i = 0; i < b.size(); i++)
    cout << b[i]->vf() << endl;
  purge(b);
} ///:~

If you can always arrange for a virtual base class to have a default constructor, you’ll make things much easier for anyone who inherits from that class.

Contents | Prev | Next

Go to CodeGuru.com
Contact: webmaster@codeguru.com
� Copyright 1997-1999 CodeGuru