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在 xna 中，当引用精灵绘图的坐标时，我有一个位置为 vector2 的精灵。（暂时忽略 z，这是基于屏幕位置正确堆叠精灵的失败尝试）。当手动引用vector2坐标时，我似乎总是在原点结束，请明白我的意思。

//draw sprite, this is called repeatedly during gameplay, it's inside a draw function
theGame.spriteBatch.Draw(headings[heading][frame], new Rectangle((int)theSprite.position.X, (int)theSprite.position.Y, theSprite.width, theSprite.height), null, Color.White, 0f, Vector2.Zero, SpriteEffects.None, theSprite.z);

在通过引用它的 X 和 Y 矢量坐标来绘制精灵时，一切都是笨拙的。如果我尝试以任何其他我能想到的方式手动使用这些相同的坐标，坐标似乎总是以原点（左上角）结束。除此之外，一切顺利，精灵的运动很好

//the movement code for the particular sprite:
// this is a function i call repeatedly during gameplay, it moves the sprite
        public void Move(GameTime Time)
        {
            if (!move) { return; } //not even moving? exit function

            direction = destination - position;//these are all vector2 orbjects
            //speed is a float
            position += direction * (speed * (float)Time.ElapsedGameTime.TotalSeconds);

            syncBounds();

            if (Vector2.Distance(position,destination)<bounds_sphere.Radius) //close enough? stop moving
            { 
                move = false;
            }
         }

//here's the moveTo function which triggers the move-event, i typically call it with screen coordinates
        public void moveTo(float _x, float _y, string _anim)

            destination = new Vector2(_x - (width / 2), _y - (height / 2));

            curr_anim = _anim;
            move = true; //kicks off trigger 
        }

 //here's an example of moveTo working with basic coordinates
 // Process touch events
        TouchCollection touchCollection = TouchPanel.GetState();
        foreach (TouchLocation tl in touchCollection)
        {
            if ((tl.State == TouchLocationState.Pressed))//|| (tl.State == TouchLocationState.Moved))
            {
                float px = tl.Position.X; //touch x
                float py = tl.Position.Y; //touch y

                gameObjects.Sprites[player].moveTo(px, py, "running");

            }
        }

所以在使用屏幕坐标时，精灵会很好地移动到目的地；但是如果我使用精灵自己的向量-x 和 y 坐标作为其目的地的一部分会发生什么，随着时间的推移，精灵最终会出现在左上角，就像命运一样。为什么会这样？它发生在手动参考坐标时，也以其他方式发生，例如碰撞检测。我实际上可以看到 2 个坐标在循环中向下倒数到原点，即使我将向量归零并将其坐标重新应用回向量也会发生这种情况（我假设除了删除任何幅度之外什么都不做，如果 vector2 甚至持有该信息）。谢谢您的帮助！

// i don't think this actually does anything
vector2 savepos = position;
position = Vector2.Zero;
position.x = savepos.x;
position.y = savepos.y;


//example of my code failing when referencing the vector2 coordinates
//sprite (me) auto-flee function code
int x = (int) me.position.X;
int y = (int) me.position.Y;
int dist = 2;
me.moveTo(rand.Next(x-dist,x+dist), rand.Next(y-dist, y+dist), running); //kick off moveTo function, begin the animation sequence

有趣的是，如果我指定另一个精灵的 vector2 位置坐标，它似乎工作正常：

//this works and the sprites move to the other object
me.moveTo(gameObjects.Sprites["anotherspriteobject"].position.X, gameObjects.Sprites["anotherspriteobject"].position.Y, running); //kick off moveTo function, begin the animation sequence
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1 回答 1

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如果你经常使用这些参数调用 moveTo 函数，精灵自然会出现在左上角。精灵的位置成员是它的左上角。但是，在 moveTo 函数中，您期望中心的位置。这就是为什么你减去它的一半大小。因此，当您将精灵的中心永久移动到其左上角时，当您阻止精灵离开时，它将导致屏幕的左上角。

解决方案是使用精灵的实际中心点作为参考：

int x = (int) me.position.X + me.width/2; 
int y = (int) me.position.Y + me.height/2; 
int dist = 2; 
me.moveTo(rand.Next(x-dist,x+dist), rand.Next(y-dist, y+dist), running); //kick off moveTo function

因此，您可以将 CenterPosition 属性添加到精灵的类。

于 2012-05-26T15:51:28.363 回答