Question

以下代码用于检查指针转换：

#include <iostream>
#include <cstdlib>

using std::cout;
using std::endl;

class A
{ 
    friend class B;
};

class B : private A { };

int main()
{
    B *b = new B;
    A *a = b;
    cout << b << endl;
    cout << a;
}

第11.2 / 4节N3797说：

A base class B of N is accessible at R, if 
— an invented public member of B would be a public member of N, or 
— R occurs in a member or friend of class N, and an invented public member of B would be a private or protected member of N, or 
— R occurs in a member or friend of a class P derived from N, and an invented public member of B would be a private or protected member of P, or 
— there exists a class S such that B is a base class of S accessible at R and S is a base class of N accessible at R

我认为A是B的可访问基类。

类型为“指向cv D的指针”的prvalue，其中D是类类型，可以是转换为“指向cv B指针”类型的prvalue，其中B是基数 D类（第10条）如果B是无法进入的（第11条）或模糊（10.2）D的基类，一个需要这个的程序转换是不正确的

为什么我收到编译时错误？

main.cpp:17:12: error: cannot cast 'B' to its private base class 'A'

demo

Answer 1

为了使friend声明能够使转化生效，它需要在友谊有效的范围内进行 - 也就是说，在B自己的代码中。 A将B声明为朋友的事实在B的定义之外无效。

基类指针转换

1 个答案: