当使用复制构造函数为持有它的变量分配一个新对象时，前一个对象是否被销毁？

Question

看看这段代码：

#include <iostream>
using namespace std;

class A {
    private:
        int _x;
        int _id;
        static int count;
    public:
        A(int x) : _x(x) {
            this->_id = A::count++;
            cout << "Object with id "      << this->_id
                 << " has been created."   << endl;
        }
        ~A() {
            cout << "Object with id "      << this->_id
                 << " has been destroyed." << endl;
        }

        int get_x(void) {
            return this->_x;
        }

        A add(A& object) {
            A tmp(this->_x + object._x);
            return tmp;
        }
};

int A::count = 1;

int main(void) { 
    A object_1(13);
    A object_2(5);
    A object_3(12);

    object_3 = object_1.add(object_2);

    cout << object_3.get_x() << endl;


    return 0;
}

这是程序的输出：

Object with id 1 has been created.
Object with id 2 has been created.
Object with id 3 has been created.
Object with id 4 has been created.
Object with id 4 has been destroyed.
18
Object with id 4 has been destroyed.
Object with id 2 has been destroyed.
Object with id 1 has been destroyed.

我不明白 ID 为 3 的对象发生了什么？它肯定是被创建的，但我看不到任何一行告诉我它曾经被摧毁过。你能告诉我这是怎么回事吗？

作为一个旁白的问题，为什么当我使用 return 0 时，析构函数工作正常，但是当我使用 exit(EXIT_SUCCESS) 时，我没有看到 Object with # has been destroyed 打印在屏幕就像从未调用过析构函数一样。

Answer 1

Is the previous object destroyed when the variable holding it gets assigned a new one using a copy constructor?

这个问题没有实际意义，因为不可能这样做。

当你运行

object_a = object_b;

这会调用赋值运算符（不是复制构造函数）。它不会创建或销毁任何对象（除非您的赋值运算符这样做）。

在这种情况下，您没有定义赋值运算符，因此使用默认运算符，它会用另一个对象的 ID（即 4）覆盖 object_3 的 ID。因此，当 object_3 被销毁时，它会打印 "Object with id 4 has been destroyed".

当使用复制构造函数为持有它的变量分配一个新对象时，前一个对象是否被销毁？

Is the previous object destroyed when the variable holding it gets assigned a new one using a copy constructor?

c++

destructor