派生类Private方法被调用 [英] Derived class Private method is getting called

问题描述

我有一个指向派生类对象的基类指针.方法foo()在基类中是公共的，但在派生类中是私有的.基类foo()是虚拟的.因此，当我从基类指针调用foo()时，Vptr Table具有派生类foo()的地址，但是在派生类中是私有的...所以它怎么被调用.

我了解运行时多态性，并且我也了解访问说明符在编译时有效，而虚拟概念在运行时有效.这样就不会有编译器错误.

我的问题是:这是一个漏洞，通过它我们可以调用Derived类的私有方法?或预期会以这种方式表现. 对此行为有任何好的解释.

非常感谢.

代码:

class A
{
public:
    virtual void foo()
    {
        std::cout << "In A";
    }
};


class B:public A
{
private:
    void foo()
    {
       std::cout << "In B ??? Its Private Method :-( ";
    }
};

int main()
{
    A* ptr = new B();
    ptr->foo();
    return 0;
}

解决方案

这是私有方法，但是由于它是虚拟的-可以调用.

n3690 11.5/1

虚拟函数的访问规则(第11条)由其声明确定，不受以下内容的影响 稍后会覆盖该功能的规则.

为什么呢?自

n3690 11.5/2

在调用点使用用于表示对象的表达式类型检查访问. 成员函数被调用(在上例中为B *). 对类中的成员函数的访问

通常是未知的(在上面的示例中为D).

I have a Base class pointer pointing to derived class object. The method foo() is public in base class but private in derived class. Base class foo() is virtual. So when i call foo() from Base class pointer, Vptr Table has the address of derived class foo(), BUT its private in Derived class ...so how is it getting called.??

I understand Run time polymorphism and i also understand that the Access specifiers work for compile time and Virtual concept works at run time. So there shall be No Compiler error.

My question is : Is this is a loop hole through which we can call private methods of Derived class ? or Its expected to behave this way. Any good Explanation for this behavior.

Thanks a lot in advance.

CODE :

class A
{
public:
    virtual void foo()
    {
        std::cout << "In A";
    }
};


class B:public A
{
private:
    void foo()
    {
       std::cout << "In B ??? Its Private Method :-( ";
    }
};

int main()
{
    A* ptr = new B();
    ptr->foo();
    return 0;
}

解决方案

It's private method, but since it's virtual - it can be called.

n3690 11.5/1

The access rules (Clause 11) for a virtual function are determined by its declaration and are not affected by the rules for a function that later overrides it.

Why this? Since

n3690 11.5/2

Access is checked at the call point using the type of the expression used to denote the object for which the member function is called (B* in the example above). The access of the member function in the class in which it was defined (D in the example above) is in general not known.

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