Question

我正在尝试创建一个具有精灵的游戏，每一秒都会产生，我尝试将其作为基础：https://github.com/libgdx/libgdx/wiki/A-simple-game 然而，当新的产生它时，它会破坏旧的，并且它们开始产生越来越快的速度。以下是相关代码：

ownCreate方法：

Array<Rectangle> chickens;
chickens = new Array<Rectangle>();
spawnChicken();

New Spawn Chicken方法：

private void spawnChicken() {


        int choice;
        choice = MathUtils.random(0, 3);
        switch (choice){

        case 0:
            chicken.x = MathUtils.random(0,1920-85);
            chicken.y = 0;          
            break;
        case 1:
            chicken.x = MathUtils.random(0,1920-85);
            chicken.y = 1080-85;    
            break;
        case 2:
            chicken.x = 0;
            chicken.y = MathUtils.random(0,1080-66);
            break;
        case 3:
            chicken.x = 1920-85;
            chicken.y = MathUtils.random(0,1080-66);
            break;
        }


    chicken.height = 66;
    chicken.width = 85;
    chickens.add(chicken);
    runningChickens ++;
    lastSpawnTime = TimeUtils.nanoTime();

渲染方法：

if(TimeUtils.nanoTime() - lastSpawnTime > 1000000000)
    spawnChicken();

Iterator<Rectangle> iter = chickens.iterator();

while(iter.hasNext()){      
    Rectangle chicken = iter.next();
    //movement

与wiki非常相似：

private Array<Rectangle> raindrops;
   private long lastDropTime; private void spawnRaindrop() {
      Rectangle raindrop = new Rectangle();
      raindrop.x = MathUtils.random(0, 800-64);
      raindrop.y = 480;
      raindrop.width = 64;
      raindrop.height = 64;
      raindrops.add(raindrop);
      lastDropTime = TimeUtils.nanoTime();
   }
 raindrops = new Array<Rectangle>();
   spawnRaindrop();
   if(TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnRaindrop();
 Iterator<Rectangle> iter = raindrops.iterator();
   while(iter.hasNext()) {
      Rectangle raindrop = iter.next();
      raindrop.y -= 200 * Gdx.graphics.getDeltaTime();
      if(raindrop.y + 64 < 0) iter.remove();
   }

Answer 1

把

Rectangle chicken = new Rectangle();

在spawnChicken()方法的开头。

<强>原因

如果您保留chicken成员，则每次调用spawnChicken()方法时都会替换该成员。
此外，每次尝试生成鸡时，都会将相同的对象添加到数组中。

控制和创建多个sprite数组Java Libgdx

1 个答案: