继承共享方法名称的接口 [英] Inherit interfaces which share a method name

问题描述

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)?

#include <cstdio>

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

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

class RealClass: public Interface1, public Interface2
{
public:
    virtual void Interface1::Name()
    {
        printf("Interface1 OK?\n");
    }
    virtual void Interface2::Name()
    {
        printf("Interface2 OK?\n");
    }
};

int main()
{
    Interface1 *p = new RealClass();
    p->Name();
    Interface2 *q = reinterpret_cast<RealClass*>(p);
    q->Name();
}   

我无法在VC8中移动定义。我发现Microsoft特定关键字__interface可以成功完成此任务，代码如下：

I failed to move the definition out in VC8. I found the Microsoft Specific Keyword __interface can do this job successfully, code below:

#include <cstdio>

__interface Interface1{
    virtual void Name() = 0;
};

__interface Interface2
{
    virtual void Name() = 0;
};

class RealClass: public Interface1,
                public Interface2
{
public:
    virtual void Interface1::Name();
    virtual void Interface2::Name();
};

void RealClass::Interface1::Name()
{
    printf("Interface1 OK?\n");
}

void RealClass::Interface2::Name()
{
    printf("Interface2 OK?\n");
}

int main()
{
    Interface1 *p = new RealClass();
    p->Name();
    Interface2 *q = reinterpret_cast<RealClass*>(p);
    q->Name();
}  

但是还有另一种方法来做这个更一般的工作在其他编译器？

but is there another way to do this something more general that will work in other compilers?

推荐答案

这个问题不经常出现。我熟悉的解决方案是由Doug McIlroy设计的，并出现在Bjarne Stroustrup的书中（在C ++的设计与演化第12.8节和 C ++编程语言 25.6）。根据设计&进化，有一个建议来优雅地处理这个特定的案例，但它被拒绝，因为这样的名称冲突不可能成为普遍足以保证单独的语言功能，不可能成为新手的日常工作。

This problem doesn't come up very often. The solution I'm familiar with was designed by Doug McIlroy and appears in Bjarne Stroustrup's books (presented in both Design & Evolution of C++ section 12.8 and The C++ Programming Language section 25.6). According to the discussion in Design & Evolution, there was a proposal to handle this specific case elegantly, but it was rejected because "such name clashes were unlikely to become common enough to warrant a separate language feature," and "not likely to become everyday work for novices."

不仅需要通过指向基类的指针调用 Name（）说出在操作导出类时想要的 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.

这篇关于继承共享方法名称的接口的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！