Question

我已经远离c ++很久了，所以很生锈，但是我应该能够做到这一点吗？我想构建对象，然后将其转换为更专业的派生类。

#pragma once

class GameObject
{
public:

    GameObject();
    virtual ~GameObject();

    int x, y;

}

class PlayerObject : Public GameObject
{

    PlayerObject();
    virtual ~PlayerObject();

    int x, y;
    string name;

};


class Game
{

    unique_ptr<GameObject> randomObject;
    unique_ptr<PlayerObject> player;


}

#########################



unique_ptr<GameObject> GameObjectManager::buildGameObject(string gameObjectId)
{

    unique_ptr<GameObject> gameObject;
    gameObject = make_unique< GameObject>();

    return gameObject;

}


##################################
int main(int argc, char *args[])

{


    unique_ptr<GameObject> rock = gameObjectManager.buildGameObject("ROCK"));
    unique_ptr<PlayerObject> playerObject = dynamic_cast<unique_ptr<PlayerObject>>(Game::gameObjectManager.buildGameObject("PLAYER"));



}

是否说要这样做动态转换。遇到这种情况，哪种更好的方法是什么？谢谢

Answer 1

这很标准，请参阅：covariant return type。

返回基类类型的C ++对象构建函数

1 个答案: