Prevent class inheritance in C++

Question

Recently one of my friend asked me how to prevent class inheritance in C++. He wanted the compilation to fail.

I was thinking about it and found 3 answers. Not sure which is the best one.

1) Private Constructor(s)

class CBase
{

public:

 static CBase* CreateInstance() 
 { 
  CBase* b1 = new CBase();
  return b1;
 }

private:

 CBase() { }
 CBase(CBase3) { }
 CBase& operator=(CBase&) { }


};

2) Using CSealed base class, private ctor & virtual inheritance

class CSealed
{

private:

 CSealed() {
 }

 friend class CBase;
};


class CBase : virtual CSealed
{

public:

 CBase() {
 }

};

3) Using a CSealed base class, protected ctor & virtual inheritance

class CSealed
{

protected:

 CSealed() {
 }

};

class CBase : virtual CSealed
{

public:

 CBase() {
 }

};

All the above methods make sure that CBase class cannot be inherited further. My Question is:

1. Which is the best method ? Any other methods available ?

2. Method 2 & 3 will not work unless the CSealed class is inherited virutally. Why is that ? Does it have anything to do with vdisp ptr ??

PS:

The above program was compiled in MS C++ compiler (Visual Studio). reference : http://www.codeguru.com/forum/archive/index.php/t-321146.html

Answer 1

As of C++11, you can add the final keyword to your class, eg

class CBase final
{
...

The main reason I can see for wanting to do this (and the reason I came looking for this question) is to mark a class as non subclassable so you can safely use a non-virtual destructor and avoid a vtable altogether.

Answer 2

You can't prevent inheritance (before C++11's final keyword) - you can only prevent instantiation of inherited classes. In other words, there is no way of preventing:

class A { ... };

class B : public A { ... };

The best you can do is prevent objects of type B from being instantiated. That being the case, I suggest you take kts's advice and document the fact that A (or whatever) is not intended to be used for inheritance, give it a non-virtual destructor, and no other virtual functions, and leave it at that.

Answer 3

You are going through contortions to prevent further subclassing. Why? Document the fact that the class isn't extensible and make the dtor non-virtual. In the spirit of c, if someone really wants to ignore the way you intended this to be used why stop them? (I never saw the point of final classes/methods in java either).

//Note: this class is not designed to be extended. (Hence the non-virtual dtor)
struct DontExtened
{
  DontExtened();
  /*NOT VIRTUAL*/
  ~DontExtened();
  ...
};

Answer 4

1) is a matter of taste. If I see it correctly, your more fancy 2nd and 3rd solutions move the error in certain circumstances from link time to compile time, which in general should be better.

2) Virtual inheritance is needed to force the responsibility to initialize the (virtual) base class to the most derived class from where the base class ctor is no longer reachable.

Answer 5

To answer your question, you can't inherit from CBase because in virtual inheritance a derived class would need to have direct access to the class from which it was inherited virtually. In this case, a class that would derive from CBase would need to have direct access to CSealed which it can't since the constructor is private.

Though I don't see the usefulness of it all (ie: stopping inheritance) you can generalize using templates (I don't think it compiles on all compilers but it does with MSVC)

template<class T>
class CSealed
{
    friend T;    // Don't do friend class T because it won't compile
    CSealed() {}
};

class CBase : private virtual CSealed<CBase>
{
};

Answer 6

If you can, I'd go for the first option (private constructor). The reason is that pretty much any experienced C++ programmer will see that at a glance and be able to recognize that you are trying to prevent subclassing.

There might be other more tricky methods to prevent subclassing, but in this case the simpler the better.

Answer 7

From C++11 onward, there is a clean solution that I am surprise not to see here. We can make the class final preventing any further inheritance.

class Foo final {};

class Bar: public Foo {}; // Fails to compile as Foo is marked as final

Answer 8

class myclass;

    class my_lock {
        friend class myclass;
    private:
        my_lock() {}
        my_lock(const my_lock&) {}
    };

    class myclass : public virtual my_lock {
        // ...
    public:
        myclass();
        myclass(char*);
        // ...
    };

    myclass m;

    class Der : public myclass { };

    Der dd;  // error Der::dd() cannot access
            // my_lock::my_lock(): private  member

I found it here to give credit. I am posting here just other people can easily access http://www.devx.com/tips/Tip/38482

Answer 9

To elaborate on Francis' answer: if class Bottom derives from class Middle, which virtually inherits from class Top, it is that most derived class (Bottom) that is responsible for constructing the virtually inherited base class (Top). Otherwise, in the multiple-inheritance/diamond-of-death scenario (where virtual inheritance is classically used), the compiler wouldn't know which of the two "middle" classes should construct the single base class. The Middle's constructor's call to the Top's constructor is therefore ignored when Middle is being constructed from Bottom:

class Top {
    public:
        Top() {}
}

class Middle: virtual public Top {
    public:
        Middle(): Top() {} // Top() is ignored if Middle constructed through Bottom()
}

class Bottom: public Middle {
    public:
        Bottom(): Middle(), Top() {}
}

So, in the the approach 2) or 3) in your question, Bottom() can't call Top() because it's inherited privately (by default, like in your code, but it's worth making it explicit) in Middle and thus is not visible in Bottom. (source)