如何在 C++ 中声明接口?

How do I setup a class that represents an interface? Is this just an abstract base class?

解决方案

To expand on the answer by bradtgmurray, you may want to make one exception to the pure virtual method list of your interface by adding a virtual destructor. This allows you to pass pointer ownership to another party without exposing the concrete derived class. The destructor doesn't have to do anything, because the interface doesn't have any concrete members. It might seem contradictory to define a function as both virtual and inline, but trust me - it isn't.

class IDemo
{
    public:
        virtual ~IDemo() {}
        virtual void OverrideMe() = 0;
};

class Parent
{
    public:
        virtual ~Parent();
};

class Child : public Parent, public IDemo
{
    public:
        virtual void OverrideMe()
        {
            //do stuff
        }
};

You don't have to include a body for the virtual destructor - it turns out some compilers have trouble optimizing an empty destructor and you're better off using the default.