當使用複製構造函數保存它的變量被賦予一個新變量時，以前的對象是否被銷燬？

Question

看看這個代碼：

#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; 
} 

下面是該程序的輸出：

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. 

我不明白髮生了什麼ID爲3到​​對象？它絕對是被創造出來的，但是我沒有看到線條告訴我它曾經被毀壞過。你能告訴我這裏發生了什麼事嗎？

順便問題，那爲什麼當我使用return 0，析構函數正常工作，但是當我使用exit(EXIT_SUCCESS)我沒有看到Object with # has been destroyed印在屏幕上，儘管析構函數永遠不會被調用。

Answer 1

當使用複製構造函數保存它的變量被賦予一個新變量時，前一個對象是否被銷燬？

這個問題是沒有意義的，因爲這是不可能的。

當您運行

object_a = object_b; 

這個調用賦值運算符（而不是拷貝構造函數）。它不會創建或銷燬任何對象（除非您的賦值操作符那樣做）。

在這種情況下，您尚未定義賦值運算符，因此使用默認值，它會用另一個對象的ID（它是4）覆蓋object_3的ID。所以當object_3被銷燬時，它會打印出「ID爲4的對象已被銷燬」。