c++ - 基类是对另一个对象的引用

Question

我刚刚遇到了一段对我来说很奇怪的代码(见下面的一个最小示例)，derived::base 是对另一个类型为 base 的对象的引用，有人可以帮我解答一下评论里的问题吗？

class base{
public:
   int a;
   int b;
};

class derived : public base{
public:
   double c;
   void run(const base & bs){
     ((base &) *this) = bs; // what does this line do? 
                            // Is derived::base now a copy of bs? 
                            // If yes, but why not write ((base) *this) = bs?
                            // if not, then derived::base is a reference to bs, 
                            // then does it mean the memory of derived::base
                            // and members of derived are no longer contiguous?
     std::cout << "a = " << a << std::endl;

   }
};

附言

@LightnessRacesinOrbit 的评论对解决问题有很大帮助，但我只能接受一个答案帖子，最好的是@WhiZTiM

Answer 1

void run(const base & bs){
     ((base &) *this) = bs; 
     std::cout << "a = " << a << std::endl;
}

上面的代码可以分解为:

void run(const base & bs){
     base& base_of_this_instance = *this; 
     base_of_this_instance = bs;
     std::cout << "a = " << a << std::endl;
}

derived 对象的内存可以布局为:

||  int a  |x|  int b  |y|  int c  ||   // <- x and y represents some hypothetical padding
||        base         |y|         ||   // <- We can slice `derived` and retrieve only base
||           derived               ||   // <- Memory consumed by derived

在您的derived::run 方法中，首先，获取对derived 的base 部分的引用，其次是 base 被分配给 bs。此赋值将调用 base 的复制赋值运算符。这意味着基础部分现在将保存 bs 中的任何内容的拷贝。