是否可能从C ++中的基类方法返回一个派生类？

问题描述

我想这样做：

class Derived;

class Base {
    virtual Derived f() = 0;
};

class Derived : public Base {
};

当然这不起作用，因为我不能返回一个不完整的类型。但是我也不能在base之前定义Derived，因为我也不能继承一个不完整的类型。我认为我可以使用模板作为一种解决方法（使用Derived作为Base的模板参数），但它似乎是一个真的丑陋的方式做事情。可能有另一种方式吗？

Of course this doesn't work since I can't return an incomplete type. But neither can I define Derived before base, since I can't inherit from an incomplete type either. I figure that I could use templates as a workaround (using Derived as a template argument to Base), but it seems a really ugly way of doing things. Might there be another way?

细化：我在写一个raytracer，每个Shape类有一个函数，它返回其边界框。但是，我把BBox作为Shape的一个子类，所以我可以想象它。这是不好的设计吗？

Elaboration: I'm writing a raytracer, and each Shape class has a function which returns its bounding box. However, I've made the BBox a subclass of Shape, so I can visualize it. Is this bad design?

推荐答案

问题中的代码没有问题。

There's nothing wrong with the code in your question. This

class Derived;

class Base {
    virtual Derived f() = 0;
};

class Derived : public Base {
    virtual Derived f() {return Derived();}
};

应编译正常。然而，'Base :: f（）'的调用者将需要看到'Derived'的定义。

should compile just fine. However, callers of 'Base::f()' will need to have seen the definition of 'Derived`.

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