CODEWITHSUNDEEP
javatimeoverflowlong-integerintervals
hxtk
hxtk

Reputation: 325

Java time interval percentage with greater precision

I have the following Java code:

public class Time {

  private Main m;
  private long st;
  private long et;

  public Time(Main m, long g){
    this.m = m;
    st = System.currentTimeMillis();
    et = st+g;
  }

  public void update() {
    long now = System.currentTimeMillis();
    float per;
    if (st >= et || now >= et) {
      m.setPer(100);
      return;
    }
    if (now <= st) {
      m.setPer(000);
      return;
    }
    per = (now - st) * 100 / (et - st);
    System.out.println(per);
    m.setPer(per);  
  }

}

which sets the percent of time out of the interval length has passed, which is used to render the same percentage of the border of a circle as the percentage of the interval.

My problem is that I require greater precision than is lent by the long type, because it is jumpy.

When I cast the variables to float or double, it overflows.

How can I get greater precision under these circumstances?

Upvotes: 0

Views: 347

Answers (2)

Bryan
Bryan

Reputation: 46

The arithmatic needs to be done as a double. You may be able to modify the update function as follows to get the result you are after. Notice this line in particular:

per = (now - st) * 100.0 / (et - st);

The code below changes the variables used to from longs.

public void update() {
  double now = (double)System.currentTimeMillis();
  double per;
  if (st >= et || now >= et) {
System.out.println ("st: " + st + ", et: " + et);
System.out.println(100);
    return;
  }
  if (now <= st) {
System.out.println(000);
    return;
  }
  per = (now - st) * 100.0 / (et - st);
  System.out.println(per);
}

Upvotes: 1

zenbeni
zenbeni

Reputation: 7213

System.nanoTime() should be what you are looking for with nanotime precision.

Upvotes: 0

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