Question

我们说我有一个整数数组

int timeouts [] = {1000 , 2000 , 3000 , 3500};

我想创建一个计时器，计数最多3.5秒，如果毫秒计数等于数组的一个元素，则调用相同的函数。有没有办法在不制作多个计时器的情况下很快完成这项工作？

Answer 1

public class ArraysFun{
    private static int[] timeouts = {1000,2000,3000,3500};
    public static void main(String[] sss){
        endlessCounter(0);
    }
    //Endless counter that calls your function
    public static void endlessCounter(int i){
        long start = System.currentTimeMillis();
        long now;
        do{
            now = System.currentTimeMillis();
        //checks to see if the time was elapsed
        }while(now - start<timeouts[i]);
        //call your function
        callFunction(i);
        //iterate through the timeouts array
        i = (i>= timeouts.length-1)? 0 : i+1;
        //call the counter again
        endlessCounter(i);
    }
    //just print which is the timeout that was waited before this call
    private static void callFunction(int i) {
        double duration = (double)timeouts[i]/1000.00;
        System.out.println("Function called after "+ duration + " seconds");
    }
}

Answer 2

对常规java.awt.Timer进行子类化。

public class VariableTimer extends Timer{

    private int [] millis;
    private counter = 0;

    public VariableTimer(int[] millis, ActionListener l) {
        super(millis[0], l);
        this.millis = millis;
    }

    @Override
    protected void fireActionPerformed(ActionEvent e) {
        super.fireActionPerformed(e);
        setDelay(millis[++counter%millis.length]);
        restart();
    }
}

没有测试过，希望它有效。

Answer 3

public class Timer {
    static final int timeouts [] = {1000 , 2000 , 3000 , 3500};

    private static int findMax(int [] array) {
       int max = Integer.MIN_VALUE;
       for (int i = 0; i < array.length; i++) {
           if (array[i] > max) {
              max = array[i];
           }
       }
       return max;
    }

    private static void checkIfContained(double number) {
        for (int i : timeouts) {
            if ((double) i == number) {
                System.out.println("SUCCESS!"); 
                // TODO
            }
        }
    }
    public static void main(String[] args) throws InterruptedException {
        int max = findMax(timeouts);
        double counter = 0.0;
        while (counter < max) {
            Thread.sleep(3500);
            counter += 3.5;
            checkIfContained(counter);
        }
    }

}

具有多个超时的Java计时器

3 个答案: