Question

我有一个GameLoop类，其中定义了ParticleSystem对象。

ParticleSystem包含一个Particle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 - 我该怎么做呢以有效的方式？

SOLUTION：

以下是Dan S帮助我提出的建议：

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

然后在类中定义它：

rm = new ResourceManager(mContext);

然后传递rm变量：

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

并且在Particle类中，我在构造函数中设置了一个String assetKey - 因此我可以通过名称引用Bitmap：

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。

Answer 1

在GameLoop的构造函数中创建您的位图并在那里保留对它们的引用，将它们设置为最终变量，以防止意外分配。无论何时创建新的粒子系统，都将它们传递给构造函数或者在适当的setter方法中设置它们。

示例：

// 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 }

只要不断传递GameLoop，直到完成为止。这可能效率不高，但是一旦你掌握了它，你也会看到这篇文章关于Dependency Injection。

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

跨类使用资源？

1 个答案: