C++:Derived + Base类实现单个接口？

Question

在C++中,是否可以使用base和派生类实现单个接口？

例如:

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

class Base
{
    public:
        virtual void BaseFunction(){}
};

class Derived : public Base, public Interface
{
    public: 
        void DerivedFunction(){}
};

void main()
{
    Derived derived;
}

这会失败,因为Derived无法实例化.就编译器而言,从未定义Interface :: BaseFunction.

到目前为止,我发现的唯一解决方案是在Derived中声明传递函数

class Derived : public Base, public Interface
{
    public: 
        void DerivedFunction(){}
        void BaseFunction(){ Base::BaseFunction(); }
};

有没有更好的解决方案？

---

编辑:如果重要,这是我使用MFC对话框的现实问题.

我有一个从CDialog派生的对话类(MyDialog可以说).由于依赖性问题,我需要创建一个抽象接口(MyDialogInterface).使用MyDialogInterface的类需要使用特定于MyDialog的方法,但还需要调用CDialog :: SetParent.我刚刚通过创建MyDialog :: SetParent并将其传递给CDialog :: SetParent来解决它,但是想知道是否有更好的方法.

Answer 1

C++没有注意到从Base继承的函数已经实现BaseFunction:该函数必须在派生自的类中显式实现Interface.这样改变:

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

class Base : public Interface
{
    public:
        virtual void BaseFunction(){}
};

class Derived : public Base
{
    public: 
        virtual void DerivedFunction(){}
};

int main()
{
    Derived derived;
}

如果您希望只能实现其中一个,则Interface可以分成两个接口:

class DerivedInterface
{
    public:
        virtual void DerivedFunction() = 0;
};

class BaseInterface
{
    public:
        virtual void BaseFunction() = 0;
};

class Base : public BaseInterface
{
    public:
        virtual void BaseFunction(){}
};

class Derived : public DerivedInterface
{
    public: 
        virtual void DerivedFunction(){}
};  

class Both : public DerivedInterface, public Base {
    public: 
        virtual void DerivedFunction(){}
};

int main()
{
    Derived derived;
    Base base;
    Both both;
}

注意:main必须返回int
注意:最好保留virtual在基础中虚拟的派生中的成员函数前面,即使它不是严格要求的.