继承共享方法名称的接口

问题描述

限时删除！！

有两个基类具有相同的函数名。我想继承它们，并且不同地驾驶每个方法。我如何使用单独的声明和定义（而不是在类定义中定义）？

There are two base classes have same function name. I want to inherit both of them, and over ride each method differently. How can I do that with separate declaration and definition (instead of defining in the class definition)?

  Name（）该解决方案增加了一些间接：

Not only do you need to call Name() through pointers to base classes, you need a way to say which Name() you want when operating on the derived class.  The solution adds some indirection:
class Interface1{
public:
    virtual void Name() = 0;
};

class Interface2{
public:
    virtual void Name() = 0;
};

class Interface1_helper : public Interface1{
public:
    virtual void I1_Name() = 0;
    void Name() override
    {
        I1_Name();
    }
};

class Interface2_helper : public Interface2{
public:
    virtual void I2_Name() = 0;
    void Name() override
    {
        I2_Name();
    }
};

class RealClass: public Interface1_helper, public Interface2_helper{
public:
    void I1_Name() override
    {
        printf("Interface1 OK?\n");
    }
    void I2_Name() override
    {
        printf("Interface2 OK?\n");
    }
};

int main()
{
    RealClass rc;
    Interface1* i1 = &rc;
    Interface2* i2 = &rc;
    i1->Name();
    i2->Name();
    rc.I1_Name();
    rc.I2_Name();
}
不太好，但决定不经常需要。
Not pretty, but the decision was it's not needed often.
                        
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