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好吧,我在 Unity 中使用 C# 工作,我创建了一些简单的函数来根据玩家的 z 位置无限期地生成一系列平台(游戏对象)。一旦游戏对象安全地看不见,我就会删除它们,并在每次玩家向前移动时调用此更新方法:

void spawnSectorGroup()
    {
        int i = numSectorsToSpawn; //get position of last sectors, when its exceeded by player then delete and spawn again
        while (i >= 0) {
            int j = Random.Range (0, sectors.Length);
            spawnSector (gamePosition.position, sectors [j]);
            i--;
        }
    }
    void checkAndUpdateSectors() 
    {
        //print ("Player pos"+playerPosition.position);
        //print ("last sector"+gamePosition.position);

        if (gamePosition.position.z - playerPosition.position.z <= 20) { //safe value ahead
            print ("theyre almost there, spawn new group");
            print (gamePosition.position.z - playerPosition.position.z);
            spawnSectorGroup ();

        }

        //destroy past ones
        GameObject[] objects = FindObjectsOfType(typeof(GameObject)) as GameObject[];
        //GameObject[] objects = GameObject.FindGameObjectsWithTag("Block");
        foreach (GameObject o in objects) {

            if (o.gameObject.tag == "Block" || o.gameObject.tag == "Cage" || o.gameObject.tag == "Animal" || o.gameObject.tag == "Enemy") {

                if (playerPosition.position.z - o.transform.position.z >= 100) { //safe aways back

                    print ("destroying past object");
                    print (o.transform.position);
                    Destroy (o.gameObject);
                }
            }

        }

    }
void spawnSector(Vector3 relativePosition,GameObject sector){
        GameObject newSector = GameObject.Instantiate (sector, relativePosition, transform.rotation) as GameObject;
        gamePosition.position += newSector.GetComponent<Sector> ().length;
    }

这一切都行得通,但是从长远来看,我担心如果每次玩家在最后一个生成的扇区的 20 个左右单位范围内生成大约 10 个新“扇区”会累积并从众多游戏对象中产生滞后。

我没有无限期产卵的经验 - 这会是个问题吗?或者这是无限期产卵的好习惯?

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1 回答 1

7

创建和销毁对象的成本很高,并且您希望避免在游戏运行时大量执行此操作。

查看有关对象池的 Unity 教程。基本思想是,不是创建和销毁对象,而是从现有对象池中获取它们,并在完成后返回它们,以便可以重用它们。

于 2016-11-04T03:43:35.097 回答