Question

我在想...有没有办法可以在特定的时间内从数字中减去32？比如500密耳？如果你能提供帮助，那就太好了！

谢谢！

public void update() {
x += dx;

if(this.y % 32 == 0) {
    this.tileY = this.y / 32;
}

if(this.x % 32 == 0) {
    this.tileX = this.x / 32;
}

System.out.println(tileX);

    }

 public void moveLeft () {
// subtract 32 dx in 500 ms
 }

Answer 1

嗯，这是我为你开发的可爱代码。我添加了关键字static，以便能够在不创建任何对象的情况下从main调用它，但它不会使用静态上下文中的任何内容。

正如我在代码中的注释试图解释的那样，这不是一个完美的解决方案，它只是一个开始，你可能会遇到诸如多线程错误之类的问题（如果你决定使用一个单独的线程来更新位置）如果方法的主体需要一段时间来执行，则会出现轻微的计时问题。

如果您认为纳秒精度对于您的目的来说有点太多，请记住还有Thread.sleep(int milis)。

以下是代码（尝试更改调用moveLeft(int, int)的值以查看结果）：

public class Slider {

  public static void main(String[] args) {
    Thread thread = new Thread() {
        @Override
        public void run() {
            /*
             * If you are going to use something like this, beware you are multi-threading
             * Make sure what you do is thread-safe
             */
            moveLeft(32, 500);
        }
    };
    thread.start();
  }

  public static void moveLeft(int distance, int milis) {
    //time_px is how many nanoseconds the Thread can sleep until it has to move 1 dx
    double time_px = (100000*milis)/distance;
    if (time_px >= 1) {
        //Get the milis and nanos, rounding for Thread.sleep
        long time_round = (long) Math.floor(time_px);
        long milis_sleep = time_round/100000;
        System.out.print("Waiting " + milis_sleep + "ms ");
        int nano_sleep = (int) (time_round%100000); 
        System.out.println(nano_sleep + "ns per dx");
        for (int i=0; i<distance; i++) {
            try {
                Thread.sleep(milis_sleep, nano_sleep);
                /*
                 * Your code here
                 * A long code here might not get you the desired result since the sleeping does
                 * not account for the time spent processing the code. But this is a good start
                 */
                System.out.println("moving 1 dx");
            } 
            catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }
    else {
        System.out.println("Cannot go that fast");
        //If you are moving that fast (more than 1 dx per nanosecond) then you need to change this up a little.
    }
  }
}

Java - 在特定时间内从数字中减去32

1 个答案: