C ++中的接口继承 [英] Interface Inheritance in C++

问题描述

我有以下类结构：

class InterfaceA
{ 
   virtual void methodA =0;
}

class ClassA : public InterfaceA
{
   void methodA();
}

class InterfaceB : public InterfaceA
{
   virtual void methodB =0;
}

class ClassAB : public ClassA, public InterfaceB
{ 
   void methodB(); 
}

现在以下代码无法编译：

Now the following code is not compilable:

int main()
{
    InterfaceB* test = new ClassAB();
    test->methodA();
}

编译器说方法 methodA（）是虚拟的，未实现。我认为它是在 ClassA 中实现的（它实现了 InterfaceA ）。
有谁知道我的错在哪里？

The compiler says that the method methodA() is virtual and not implemented. I thought that it is implemented in ClassA (which implements the InterfaceA). Does anyone know where my fault is?

推荐答案

那是因为你有两份了InterfaceA 。有关更大的解释，请参阅此处： https://isocpp.org/wiki/faq/multiple-inheritance （你的情况类似于'可怕的钻石'）。

That is because you have two copies of InterfaceA. See this for a bigger explanation: https://isocpp.org/wiki/faq/multiple-inheritance (your situation is similar to 'the dreaded diamond').

你需要添加关键字 virtual 从InterfaceA继承ClassA时。从InterfaceA继承InterfaceB时，还需要添加 virtual 。

You need to add the keyword virtual when you inherit ClassA from InterfaceA. You also need to add virtual when you inherit InterfaceB from InterfaceA.

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