关于c ++：派生类中的私有虚函数

Private virtual function in derived class

本问题已经有最佳答案，请猛点这里访问。

Possible Duplicate:
C++: overriding public\private inheritance

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class base {
public:
virtual void doSomething() = 0;
};

class derived : public base {
private: // <-- Note this is private

virtual void doSomething()
{ cout <<"Derived fn" << endl; }
};

现在，如果我的以下的：

1 2	base *b = new child; b->doSomething(); // Calls the derived class function even though that is private

问题：

它是能够调用的函数衍生类虽然它是私人的。这是如何可能的？

现在，如果我改变继承的访问说明符从公共到私人保护/编译错误：I get a

1	'type cast' : conversion from 'Derived ' to 'base ' exists, but is inaccessible

注：我知道的概念specifiers继承的访问。第二个案例的操作系统在它的源的保护，它的inaccessible。但我想先了解到问题的答案。任何输入将是非常理解的。

在第一种情况下，对调用所通过的静态类型执行访问检查(总是这样)。*b的静态类型是base，在这种情况下，doSomething()是public。

C++ 03 11.6"访问虚拟函数"说：

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. [Example:

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class B {
public:
virtual int f();
};

class D : public B {
private:
int f();
};

void f()
{
D d;
B* pb = &d;
D* pd = &d;

pb->f(); //OK:B::f()is public,
// D::f() is invoked
pd->f(); //error:D::f()is private
}

—end example]

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.

号

特别要记住，"成员函数在定义它的类中的访问(在上面的例子中是d)通常是未知的"。一般来说，在您的示例中，当调用b->doSomething();时，编译器可能根本不了解derived或child，更不用说访问derived::doSomething()是私有的。

访问控制是在编译时实现的，而不是运行时，而多态性(包括使用虚拟函数)是运行时特性。