CODEWITHSUNDEEP
javadoubleintrounding
David
David

Reputation: 14953

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

Right now I'm trying this: 

int a = round(n);

where n is a double but it's not working. What am I doing wrong?

Upvotes: 138

Views: 332119

Answers (8)

Divyansh Rai
Divyansh Rai

Reputation: 190

The Math.round function is overloaded When it receives a float value, it will give you an int. For example this would work.

int a=Math.round(1.7f);

When it receives a double value, it will give you a long, therefore you have to typecast it to int.

int a=(int)Math.round(1.7);

This is done to prevent loss of precision. Your double value is 64bit, but then your int variable can only store 32bit so it just converts it to long, which is 64bit but you can typecast it to 32bit as explained above.

Upvotes: 9

Alexander Cyberman
Alexander Cyberman

Reputation: 2168

public static int round(double d) {
    if (d > 0) {
        return (int) (d + 0.5);
    } else {
        return (int) (d - 0.5);
    }
}

Upvotes: 2

thug-gamer
thug-gamer

Reputation: 355

Documentation of Math.round says:

Returns the result of rounding the argument to an integer. The result is equivalent to (int) Math.floor(f+0.5).

No need to cast to int. Maybe it was changed from the past.

Upvotes: 1

Dr. Daniel Thommes
Dr. Daniel Thommes

Reputation: 697

If you don't like Math.round() you can use this simple approach as well:

int a = (int) (doubleVar + 0.5);

Upvotes: 30

Joey Gibson
Joey Gibson

Reputation: 7104

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.

Upvotes: -1

Mihir Mathuria
Mihir Mathuria

Reputation: 6539

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) Math.round(doubleVar);

Upvotes: 276

valerybodak
valerybodak

Reputation: 4413

Rounding double to the "nearest" integer like this:

1.4 -> 1

1.6 -> 2

-2.1 -> -2

-1.3 -> -1

-1.5 -> -2

private int round(double d){
    double dAbs = Math.abs(d);
    int i = (int) dAbs;
    double result = dAbs - (double) i;
    if(result<0.5){
        return d<0 ? -i : i;            
    }else{
        return d<0 ? -(i+1) : i+1;          
    }
}

You can change condition (result<0.5) as you prefer.

Upvotes: 12

Scott
Scott

Reputation: 4200

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();
  }
}

Upvotes: 3

Related Questions