c++ - 为什么我调用构造函数和析构函数的顺序很奇怪

Question

我正在研究构造函数和析构函数的调用顺序，我编写了该代码：

#include <string>
#include <vector>
#include <iostream>

using namespace std;

class Corgi {

    private:
        string nickname;

    public:
        Corgi(): nickname("unknown") {
            cout << "Corgi object CREATED. Name: unknown" << endl;
        }
        Corgi(const string& _nickname): nickname(_nickname) {
            cout << "Corgi object CREATED.   \tName: " + nickname << endl;
        }
        string getName() { return nickname; }
        ~Corgi() { cout << "Corgi object DESTROYED. \tName: " << nickname << endl; }
};

int main() {
vector<Corgi> zoo;
    for (int i = 0; i < 5; i++)
        zoo.push_back(Corgi("Zoo Corgi_" + to_string(i)));

    return 0;
}

我得到的输出：

Corgi object CREATED.       Name: Zoo Corgi_0
Corgi object DESTROYED.     Name: Zoo Corgi_0
Corgi object CREATED.       Name: Zoo Corgi_1
Corgi object DESTROYED.     Name: Zoo Corgi_0
Corgi object DESTROYED.     Name: Zoo Corgi_1
Corgi object CREATED.       Name: Zoo Corgi_2
Corgi object DESTROYED.     Name: Zoo Corgi_0
Corgi object DESTROYED.     Name: Zoo Corgi_1
Corgi object DESTROYED.     Name: Zoo Corgi_2
Corgi object CREATED.       Name: Zoo Corgi_3
Corgi object DESTROYED.     Name: Zoo Corgi_3
Corgi object CREATED.       Name: Zoo Corgi_4
Corgi object DESTROYED.     Name: Zoo Corgi_0
Corgi object DESTROYED.     Name: Zoo Corgi_1
Corgi object DESTROYED.     Name: Zoo Corgi_2
Corgi object DESTROYED.     Name: Zoo Corgi_3
Corgi object DESTROYED.     Name: Zoo Corgi_4
Corgi object DESTROYED.     Name: Zoo Corgi_0
Corgi object DESTROYED.     Name: Zoo Corgi_1
Corgi object DESTROYED.     Name: Zoo Corgi_2
Corgi object DESTROYED.     Name: Zoo Corgi_3
Corgi object DESTROYED.     Name: Zoo Corgi_4

这个输出让我完全困惑。

• 如果 Zoo Corgi_0 只创建一次，为什么它会被销毁 5 次？
• 为什么 Zoo Corgi_0 在 Zoo Corgi_1 被创建之前就被创建并立即销毁？
• 是什么让 C++ 在创建 Zoo Corgi_1 后再次破坏 Zoo Corgi_0？

在输出中，我预计有 10 行：5 行用于从 0 到 4 创建 Zoo Corgis，5 行用于破坏这些对象。但显然事情要复杂得多。

我很感激任何帮助。

标签： c++constructordestructor