为什么我可以通过指向派生对象的基类指针访问派生私有成员函数?

Why can I access a derived private member function via a base class pointer to a derived object?(为什么我可以通过指向派生对象的基类指针访问派生私有成员函数?)

问题描述

#include<iostream>

using namespace std;
class base
{
public:
    virtual void add() {
        cout << "hi";
    }
};

class derived : public base
{
private:
    void add() {
        cout << "bye";
    }
};

int main()
{
    base *ptr;
    ptr = new derived;
    ptr->add();
    return 0;
}

输出是再见

我对如何实现没有问题.我了解您使用 vtables，并且 derived 的 vtable 包含新的 add() 函数的地址.但是 add() 是私有的，当我尝试在类外访问它时，编译器不应该产生错误吗?不知怎的，这似乎不对.

I dont have a problem with how this is implemented. I understand you use vtables and the vtable of derived contains the address of the new add() function. But add() is private shouldn't compiler generate an error when I try to access it outside the class? Somehow it doesn't seem right.

推荐答案

add()在derived中只是私有的，而静态类型 你拥有的是 base* - 因此 base 的访问限制适用.
通常，您甚至无法在编译时知道指向 base 的指针的动态类型是什么，例如根据用户输入进行更改.

add() is only private in derived, but the static type you have is base* - thus the access restrictions of base apply.
In general you can't even know at compile time what the dynamic type of a pointer to base will be, it could e.g. change based on user input.

这是根据 C++03 §11.6:

虚函数的访问规则(第 11 条)由其声明决定，不受稍后覆盖它的函数规则的影响.
[...]在调用点使用表达式类型检查访问，该表达式用于表示调用成员函数的对象[...].对定义它的类中的成员函数的访问 [...] 通常是未知的.

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.
[...] 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 [...]. The access of the member function in the class in which it was defined [...] is in general not known.

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