Rounding a double to turn it into an int (java)

Question

Right now i'm trying this:

    int a = round(n) ; 

where n is a double but its not working. What am i doing wrong?

Answer 1

import java.math.*;
public class TestRound11 {
  public static void main(String args[]){
    double d = 3.1537;
    BigDecimal bd = new BigDecimal(d);
    bd = bd.setScale(2,BigDecimal.ROUND_HALF_UP);
    // output is 3.15
    System.out.println(d + " : " + round(d, 2));
    // output is 3.154
    System.out.println(d + " : " + round(d, 3));
  }

  public static double round(double d, int decimalPlace){
    // see the Javadoc about why we use a String in the constructor
    // http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html#BigDecimal(double)
    BigDecimal bd = new BigDecimal(Double.toString(d));
    bd = bd.setScale(decimalPlace,BigDecimal.ROUND_HALF_UP);
    return bd.doubleValue();
  }
}

Answer 2

int a = Math.round(n);

Answer 3

You really need to post a more complete example, so we can see what you're trying to do. From what you have posted, here's what I can see. First, there is no built-in round() method. You need to either call Math.round(n), or statically import Math.round, and then call it like you have.

Answer 4

What is the return type of the round() method in the snippet?

If this is the Math.round() method, it returns a Long when the input param is Double.

So, you will have to cast the return value:

int a = (int) round(doubleVar)