c++ - 为什么显式类型转换允许向上转换为私有(private)继承？

Question

#include<iostream>
using namespace std;

class A {
    public:
    void f(){cout<<"A"<<endl;}
};

class B : private A {
    public:
    void f(){cout<<"B"<<endl;}
};

int main (){

由于 B 类私下继承 A 类，因此这种向上转换不应该起作用:

    A* a = new B;

但是显式类型转换是允许的。为什么？

    A* a1 = (A*)new B;
    a1->f();
    return 0;
}

Answer 1

类型转换

A* a1 = (A*)new B;

是对不可访问的基类的强制转换。

它只能表示为 C 风格的类型转换。如果在这种情况下可以使用 static_cast，则它等效于 static_cast 的作用，而不等效于 reinterpret_cast。特别是结果地址不一定与参数地址相同。

C++11 §5.4/4:

” The same semantic restrictions and behaviors [as for a static_cast] apply [for a C style cast], with the exception that in performing a static_cast in the following situations the conversion is valid even if the base class is inaccessible:

— a pointer to an object of derived class type or an lvalue or rvalue of derived class type may be explicitly converted to a pointer or reference to an unambiguous base class type, respectively;