C++ abstract class without pure virtual functions?

问题

I have a base class

class ShapeF
{
public:
    ShapeF();
    virtual ~ShapeF();

    inline void SetPosition(const Vector2& inPosition) { mPosition.Set(inPosition); }

protected:
    Vector2 mPosition;
}

Obviously with some ommitied code, but you get the point. I use this as a template, and with some fun (ommited) enums, a way to determine what kind of shape i'm using

class RotatedRectangleF : public ShapeF
{
public:
    RotatedRectangleF();
    virtual ~RotatedRectangleF();
protected:
    float mWidth;
    float mHeight;
    float mRotation;
}

ShapeF does its job with the positioning, and an enum that defines what the type is. It has accessors and mutators, but no methods.

Can I make ShapeF an abstract class, to ensure nobody tries and instantiate an object of type ShapeF?

Normally, this is doable by having a pure virtual function within ShapeF

//ShapeF.h
virtual void Collides(const ShapeF& inShape) = 0;

However, I am currently dealing with collisions in a seperate class. I can move everything over, but i'm wondering if there is a way to make a class abstract.. without the pure virtual functions.

回答1:

You could declare, and implement, a pure virtual destructor:

class ShapeF
{
public:
    virtual ~ShapeF() = 0;
    ...
};

ShapeF::~ShapeF() {}

It's a tiny step from what you already have, and will prevent ShapeF from being instantiated directly. The derived classes won't need to change.

回答2:

Try using a protected constructor

回答3:

If your compiler is Visual C++ then there is also an "abstract" keyword:

class MyClass abstract
{
    // whatever...
};

Though AFAIK it will not compile on other compilers, it's one of Microsoft custom keywords.

来源：https://stackoverflow.com/questions/14631685/c-abstract-class-without-pure-virtual-functions

标签

c++

polymorphism

abstract-class

pure-virtual