C ++析构函数未调用

我有三节课：房间，门和世界

#include <set>;

using namespace std;

class Door; // Forward declaration

class Room {
public:
    Door* door1;
    Door* door2;

    Room(){}

    ~Room() {
        delete door1;
        door1 = 0;
        delete door2;
        door2 = 0;
    }
};

class Door {
public:
    Room* roomA;
    Room* roomB;

    Door(Room* roomA, Room* roomB) {
        this->roomA = roomA;
        this->roomB = roomB;
        linkRooms(); // This sets up the Door-Pointers in the Rooms
                     // so they know about the new door.
    }    

    ~Door() {
        // Returns the room-owned pointer pointing at this door
        getMyRoomPointer(roomA) = 0;
        getMyRoomPointer(roomB) = 0;
    }    

    Door * & getMyRoomPointer(Room * const & room) {
        if (room->door1 == this) return room->door1;
        else return room->door2;
    }

    void linkRooms() {
        roomA->door1 = this;
        roomB->door2 = this;
    }

};

class World {
public:
    std::set<Room*> rooms;

    World() {
        // Set up two rooms and link them using a door
        Room* newRoom = new Room();
        rooms.insert(newRoom);
        Room* anotherNewRoom = new Room();
        rooms.insert(anotherNewRoom);

        new Door(newRoom, anotherNewRoom);
    }
    ~World() {
        // Iterate over the rooms and call delete on all of them
        for (std::set<Room*>::iterator it = rooms.begin(); it != rooms.end(); ++it) {
            delete *it;
        }
    }
};

int main() {

    World world;

    return 0;
}

在运行main时，构造函数仅用两个房间和一个门作为它们之间的链接来填充整个世界。主要返回后，应删除世界，并应注意所有房间和门。

事实是，我的Door析构函数从未被调用过。因此，房间内的“门”指针未设置为null，而“另一侧”的房间试图删除同一扇门时出现错误。

当我只创建Door的实例，然后立即删除它时，我没有遇到任何问题：

int main(){

    Room oneRoom;
    Room anotherRoom;

    Door* door = new Door(&oneRoom, &anotherRoom);
    delete door; // works just fine

    return 0;
}

问题：为什么不调用Door构造函数？是否可以像第一个示例一样设置“门”？

我知道，我要双重删除房间的门指针，并且可以（并且应该）使用SmartPointers。现在，我只是想知道为什么我要面对这种行为。毕竟，我还是C ++的新手。

我现在确实设置了一个可运行的示例，该示例再现了该错误。