Question

我遇到了问题。要制作视频游戏，我必须实现一个名为update（）的函数。

参数lastCallTime是我们最后一次调用update。

必须执行以下操作：

计算自上次调用更新以来经过的时间。
如果经过的时间超过20 [ms]而不是ifUpdate = true。
并重置每个对象的lastUpdated时间。

private long lastUpdated;               //The last updated time for this object.

public boolean update (long lastCallTime){
    boolean ifUpdate = false;

    long currentTime = System.currentTimeMillis();
    long delta = (currentTime - lastCallTime);
    if (delta >= 20){
        //Do something and
        ifUpdate = true;
        lastUpdated = 0;
    }
    return ifUpdate;
}

线程将调用此方法：

@Override
public void run(){
    long lastCallTime;
    int k = 0;

    lastCallTime = System.currentTimeMillis();
    while(true){
        //Do something
        boolean ifUpdate = update(lastCallTime);          
    }
}

Answer 1

前段时间我写了类似的东西。更新方法将以FPS速率调用（示例中为24）。因此，需要计算更新方法所消耗的时间（变量timeDiff）。此外，还检查过程是否睡眠多于或少于指示（变量overSleepTime），并且此时间将在下一个循环中添加/减去。

我希望它可以帮到你

long timeDiff, sleepTime;
long afterTime, beforeTime;
long overSleepTime = 0L;

int fps = 24; // the desire FPS
long period = 1000000000L / fps;

while(true) {
  beforeTime = System.nanoTime();

  // your game logic
  update();

  afterTime = System.nanoTime();

  // timeDiff is time needed by the update process
  timeDiff = afterTime - beforeTime;
  sleepTime = (period - timeDiff) - overSleepTime;

  if(sleepTime > 0) {
    try {
      Thread.sleep(sleepTime / 1000000L); // sleeptTime is in nano secs and sleep expects millis
    } catch(InterruptedException e) {}

    // checks if the thread has slept more than desired
    overSleepTime = (System.nanoTime() - afterTime) - sleepTime;

  } else {
    // negative sleeptTime means that the system can't be update at the desired rate
    overSleepTime = 0L;
  }
}

}

每x毫秒更新一个对象

1 个答案: