请看一下我的 pong 游戏的以下结构。
gameLoop();方法
//Only run this in another Thread!
private void gameLoop()
{
//This value would probably be stored elsewhere.
final double GAME_HERTZ = 30.0;
//Calculate how many ns each frame should take for our target game hertz.
final double TIME_BETWEEN_UPDATES = 1000000000 / GAME_HERTZ;
//At the very most we will update the game this many times before a new render.
//If you're worried about visual hitches more than perfect timing, set this to 1.
final int MAX_UPDATES_BEFORE_RENDER = 5;
//We will need the last update time.
double lastUpdateTime = System.nanoTime();
//Store the last time we rendered.
double lastRenderTime = System.nanoTime();
//If we are able to get as high as this FPS, don't render again.
final double TARGET_FPS = 60;
final double TARGET_TIME_BETWEEN_RENDERS = 1000000000 / TARGET_FPS;
//Simple way of finding FPS.
int lastSecondTime = (int) (lastUpdateTime / 1000000000);
while (running)
{
double now = System.nanoTime();
int updateCount = 0;
if (!paused)
{
//Do as many game updates as we need to, potentially playing catchup.
while( now - lastUpdateTime > TIME_BETWEEN_UPDATES && updateCount < MAX_UPDATES_BEFORE_RENDER )
{
updateGame();
lastUpdateTime += TIME_BETWEEN_UPDATES;
updateCount++;
}
//If for some reason an update takes forever, we don't want to do an insane number of catchups.
//If you were doing some sort of game that needed to keep EXACT time, you would get rid of this.
if ( now - lastUpdateTime > TIME_BETWEEN_UPDATES)
{
lastUpdateTime = now - TIME_BETWEEN_UPDATES;
}
//Render. To do so, we need to calculate interpolation for a smooth render.
float interpolation = Math.min(1.0f, (float) ((now - lastUpdateTime) / TIME_BETWEEN_UPDATES) );
//float interpolation = 1.0f;
drawGame(interpolation);
lastRenderTime = now;
//Yield until it has been at least the target time between renders. This saves the CPU from hogging.
while ( now - lastRenderTime < TARGET_TIME_BETWEEN_RENDERS && now - lastUpdateTime < TIME_BETWEEN_UPDATES)
{
Thread.yield();
//This stops the app from consuming all your CPU. It makes this slightly less accurate, but is worth it.
//You can remove this line and it will still work (better), your CPU just climbs on certain OSes.
//FYI on some OS's this can cause pretty bad stuttering. Scroll down and have a look at different peoples' solutions to this.
try {Thread.sleep(1);} catch(Exception e) {}
now = System.nanoTime();
}
}
}
}
updateGame();方法
if(p1_up){
if(player.equals("p1")){
p1.moveUp();
}
else
{
p2.moveUp();
}
}
else if(p1_down){
if(player.equals("p1")){
p1.moveDown();
}
else
{
p2.moveDown();
}
}
向上移动();向下移动()；桨法
public void moveUp(){
last_y = y;
last_x = x;
y -= 50.0;
}
public void moveDown(){
last_y = y;
last_x = x;
y += 50.0;
}
drawGame(插值);方法
public void paintComponent(Graphics g)
{
super.paintComponent(g);
for(int i=0;i<balls.size();i++){
paintBall(g, balls.get(i));
}
drawPaddle(g, p1);
drawPaddle(g, p2);
}
public void drawPaddle(Graphics g, Paddle p){
paddle_drawX = (int)((p.x - p.last_x)*interpolation + p.last_x);
paddle_drawY = (int)((p.y - p.last_y)*interpolation + p.last_y);
g.drawRect(paddle_drawX, paddle_drawY, 10, 50);
}
我是游戏编程的初学者，所以我对游戏循环不太了解。我在互联网上找到了上述固定时间步长的游戏循环，并将其用作我的游戏的游戏循环。环使球平稳移动，但 Racket 在移动时不会停留在一处。当我按下一个按键来移动桨时，桨会不断晃动 不在一处停留。桨的 y 坐标不断变化，就像
33、45、20、59、34、59、34、59、33、59、34、58
我知道问题出在插值上，因为它不断变化的值会改变渲染中桨的 y 坐标。我已经思考这个问题有一段时间了，但我不知道如何使游戏循环适用于任何 Action ，所以我来这里寻求帮助。我感谢任何建议/帮助!
这是我完整的桨类(class)。
public class Paddle
{
float x;
float y;
float last_y;
float last_x;
public Paddle(int x, int y)
{
this.x = x;
this.y = y;
this.last_x = x;
this.last_y = y;
}
public void setNewX(int d){
last_y = y;
last_x = x;
x = d;
}
public void setNewY(int d){
last_y = y;
last_x = x;
y = d;
}
public void moveUp(){
last_y = y;
last_x = x;
y -= 50.0;
}
public void moveDown(){
last_y = y;
last_x = x;
y += 50.0;
}
}
我通过全局变量在主类中启动桨位置。
public Paddle p1 = new Paddle(10, 10);
public Paddle p2 = new Paddle(950, 10);
我有以下事件监听器来处理击键。
Action handle_up_action = new AbstractAction(){
public void actionPerformed(ActionEvent e){
p1_up = true;
}
};
Action handle_up_action_released = new AbstractAction(){
public void actionPerformed(ActionEvent e){
p1_up = false;
}
};
Action handle_down_action = new AbstractAction(){
public void actionPerformed(ActionEvent e){
p1_down = true;
}
};
Action handle_down_action_released = new AbstractAction(){
public void actionPerformed(ActionEvent e){
p1_down = false;
}
};
最佳答案
您想通过 interpolation 实现什么目标？根据我的理解，它代表上一个和下一个“更新时间”之间耗时百分比。
因此它应该每 33.3 毫秒从 0 连续进展到 1。
不知道你怎么用这个interpolation paintBall 中的变量方法，但对于桨，它会将你的桨绘制在 p.x;p.y 之间的“伪随机位置”和p.last_x;p.last_y (取决于两者之间的时间updateGame())。
为了纠正这个问题，从循环逻辑中，您应该了解每个游戏实体(球、桨等)必须有两种状态(位置):
- 逻辑状态，仅在每次 TIME_BETWEEN_UPDATES 时更新
- 视觉状态，可以在每次渲染时随时更新。
这就像您有一组点(代表逻辑状态)并且您想要在这些点(代表视觉状态)之间的任意位置进行插值。 你的代码就像 this .
第一个解决方案
纠正桨抖动的最简单方法是避免插值和使用:
public void drawPaddle(Graphics g, Paddle p){
paddle_drawX = (int)p.x;
paddle_drawY = (int)p.y;
g.drawRect(paddle_drawX, paddle_drawY, 10, 50);
}
但是你的 Action 看起来像 this (视觉位置只会改变每个 TIME_BETWEEN_UPDATES )
第二种解决方案
您想要p.x;p.y作为逻辑位置，但视觉位置应插入 p.last_x;p.last_y 之间以及逻辑位置(如果渲染是在输入处理和下一个 updateGame() 之间完成的):您必须重置 p.last_x;p.last_y当updateGame()叫做。要实现此目的，请在 updateGame() 内调用桨的 updateMovement() 方法.
public void updateMovement(){
last_y = y;
last_x = x;
}
您可以有其他解决方案，例如使用速度变量或移动函数，以获得平滑的移动、加速度等。它主要是第二种解决方案的推广。它需要更大的改变，但更灵活、更强大。为了实现这一点，您可能需要在桨中存储最后的“更新位置”以及所有与运动相关的变量，例如运动开始日期。添加一个方法来检索可以使用两次更新之间的任何日期调用的“视觉位置”，以及一个更新“逻辑位置”的方法，调用每个 updateGame() .
关于java - 乒乓 Racket 不停地晃动，不停留在一个位置上，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20365539/