C ++：使用基类作为接口的实现 [英] C++: using a base class as the implementation of an interface

问题描述

在C ++中，是否可以使用另一个基类来提供派生类中的接口（即抽象基类）的实现？

In C++ is it possible to use another base class to provide the implementation of an interface (i.e. abstract base class) in a derived class?

class Base
{
    virtual void myfunction() {/*...*/};
}
class Interface
{
    virtual void myfunction() = 0;
}
class Derived
    : public Base, public Interface
{
    // myfunction is implemented by base
}

上面应该编译，但实际上不工作。有什么办法实现这种效果吗？

The above should compile, but doesn't actually work. Is there any way to achieve this effect?

如果任何人关心，想要这一点的原因是（使我的应用程序）使用通用库另一个项目/命名空间提供在我的项目中实现一个接口。我可以只是包装一切，但这似乎很多额外的开销。

In case anyone cares, the reason for wanting this is that it make sense (for my application) to use a generic library from another project/namespace to provide the implementation of an interface in my project. I could just wrap everything, but that seems like a lot of extra overhead.

谢谢。

推荐答案

如果 Base 不是从接口派生的，在派生中转发呼叫。它只是开销的意义上，你必须编写额外的代码。我怀疑优化器会像你原来的想法一样有效。

If Base isn't derived from Interface, then you'll have to have forwarding calls in Derived. It's only "overhead" in the sense that you have to write extra code. I suspect the optimizer will make it as efficient as if your original idea had worked.

class Interface {
    public:
        virtual void myfunction() = 0;
};

class Base {
    public:
        virtual void myfunction() {/*...*/}
};

class Derived : public Interface, public Base {
    public:
        void myfunction() { Base::myfunction(); }  // forwarding call
};

int main() {
   Derived d;
   d.myfunction();
   return 0;
}

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