c++ - 离开范围时析构函数调用的顺序？ (C++)

Question

我试图了解退出范围时析构函数调用的顺序。假设我有以下代码:

class Parent{

Parent(){cout<<"parent c called \n";}
~Parent(){cout<< "parent d called \n";}
};

class Child: public parent{

Child(){cout<< "child c called \n";}
~Child(){cout<<"child d called\n";}
};

现在，我知道子构造函数和析构函数是从父派生的，所以下面是主要的:

int main(){

Parent Man;
Child Boy;

return 0;
}

将产生输出:

parent c called
parent c called
child c called
... //Now what?

但是现在，当我退出范围时会发生什么？我有多个东西需要销毁，那么编译器如何选择顺序呢？我可以有两种输出可能性:

parent c called           |         parent c called      
parent c called           |         parent c called
child c called            |         child c called
child d called            |         parent d called
parent d called           |         child d called
parent d called           |         parent d called

如果 Boy 首先被摧毁，则适用左侧情况，如果 Man 首先被摧毁，则适用右侧情况。计算机如何决定先删除哪一个？

Answer 1

派生析构函数在祖先析构函数之前被调用。所以 Child 析构函数体将首先被调用，然后是 Parent 析构函数体。并且构造的对象以相反的顺序销毁，因此 Boy 对象将在 Man 对象被销毁之前被销毁。