为什么我复制对象时会丢弃vptr？

Question

实施例

#include <stdio.h> 
#include <stdlib.h> 
#include <iostream> 
#include <iomanip> 

struct father 
{ 
    int variable; 
    father(){variable=0xEEEEEEEE;}; 
    virtual void sing(){printf("trollolo,%x\n",variable);} 
    ~father(){}; 
}; 
struct son:father 
{ 
    son(){variable=0xDDDDDDDD;}; 
    virtual void sing(){printf("trillili,%x\n",variable);} 
    ~son(){}; 
}; 
int main() 
{ 
    father * ifather=new(father); 
    son * ison=new(son); 
    father uncle; 
    father * iteachers; 

    *((long long*)&uncle)=0xDEAF; 
    iteachers=(father*)malloc(20*sizeof(father)); 

    //ineffective assignments 
    iteachers[0]=*ifather; 
    uncle=*ifather; 

    ifather->sing();//called to prevent optimization 
    ison->sing();//only to prevent optimization 

    std::cout.setf(std::ios::hex); 
    std::cout<<"father:"<<*((long long*)ifather)<<","<<std::endl; 
    std::cout<<"teacher0:"<<*((long long*)&(iteachers[0]))<<","<<std::endl; 
    std::cout<<"uncle:"<<*((long long*)&uncle)<<","<<std::endl; 
    std::cout<<"(son:"<<*((long long*)ison)<<"),"<<std::endl; 

// uncle.sing();//would crash 
} 

教师V表指针[0]是零当用gcc编译。 此外，叔叔的vtable指针保持其原始值，而不是被覆盖。 我的问题：为什么这样呢？ 是否有一个清洁的解决方法？我可以去uncle._vptr=ifather->_vptr，仍然是便携式？什么是ORDINARY例程复制对象？我应该甚至提交一个错误？ 注意：它应该复制整个对象的平台无关，因为不管对象类型的标识如何完成，因为它应该始终在对象的数据块内！

文章

Why does my C++ object loses its VPTr

没有帮助我，必须有不同的原因。

Answer 1

据我了解，基本的问题是这样的代码：

#include <iostream> 
using namespace std; 

struct Base 
{ 
    virtual void sing() { cout << "Base!" << endl; } 
    virtual ~Base() {} 
}; 

struct Derived: Base 
{ 
    void sing() override { cout << "Derived!" << endl; } 
}; 

auto main() 
    -> int 
{ 
    Base* p = new Derived(); 
    *p = Base(); 
    p->sing();  // Reporting "Base" or "Derived"? 
} 

应报告“基地”或“派生”。

总之，赋值不会改变对象的类型。

因此，它报告“派生”。