我有一个其中定义了ParticleSystem对象的GameLoop类。

粒子系统包含一个粒子对象数组-我不想为每个单独的粒子加载图像,ID就像只能从GameLoop类中的源图像中绘制一样-如何高效地执行此操作方式?

解:

这是Dan S帮助我提出的:

public class ResourceManager  {

Context mContext;

public final HashMap<String, Bitmap> Texture = new HashMap<String, Bitmap>();

//Constructor
public ResourceManager(Context context)
{
    mContext = context;
    Texture.put("particle1", loadBitmap(R.drawable.particle1));
}

public Bitmap getBitmap(String key)
{
return Texture.get(key);
}

 /** Load an image */
    protected Bitmap loadBitmap(int id) { return   BitmapFactory.decodeResource(mContext.getResources(), id); }

  } //End ResourceManager


然后在一个类中定义它:

rm = new ResourceManager(mContext);


然后向下传递rm变量:

    ParticleSystem.Draw(rm, canvas);
   {
     Particle.Draw(rm, canvas);
    }


在“粒子”类中,我在构造函数中设置了一个字符串assetKey,以便可以通过名称引用位图:

public void doDraw(ResourceManager rm, Canvas canvas) {
    canvas.drawBitmap(rm.getBitmap(AssetKey), xpos, ypos, null);
}


我希望这对其他人也有帮助。

最佳答案

在GameLoop的构造函数中,创建位图并在其中保留对其的引用,将其设置为最终变量,以防止意外分配。每当您创建新的粒子系统时,将它们传递给构造函数或以适当的setter方法进行设置。

例:

// in GameLoop definition

private Bitmap particleA;
private Bitmap particleB;

// somewhere in GameLoop constructor
particleA = BitmapFactor.decodeResource (activity.getResources(), R.drawable.particle_a);
particleB = BitmapFactor.decodeResource (activity.getResources(), R.drawable.particle_b);

// where you build your particle
Particle (GameLoop gl, ...) {
   oneLevelDown = new OneLevelDown(GameLoop gl, ...);
}

OneLevelDown (GameLoop gl, ...) {
   twoLevelDown = new TwoLevelDown(GameLoop gl, ...);
}

TwoLevelDown (GameLoop gl, ...) {
   particleABitmap = gl.getParticleA(); // simple getter
   particleBBitmap = gl.getParticleB(); // simple getter
}


只要继续向下传递GameLoop,直到完成。这可能并不高效,但是一旦您掌握了窍门,也可以参阅有关Dependency Injection的本文。

Java内存使用示例:

class Car{
   public int year;
   public int name;
}

// Object variables are references to an object's data.

Car a = new Car(); // first car object and variable.
a.year = 1990;
a.name = "Sam";

Car b = new Car();// second car object and variable
b.year = 2000;
b.name = "Bob";

Car c = a; // third car **variable** but same car object as **a**

// primitives are different

int a = 23; // first variable, 4 bytes used
int b = 45; // second variable, 4 bytes used (and a total of 8)
int c = a;  // third variable, 4 bytes used (and a total of 12). c took on the value of a, but does not point to the same memory as a

09-25 17:54