How do I get the part after the decimal point in Java?

Question

I have a double variable d = 1.15.

I want the number after the decimal point, i.e. "15".

What is best way to achieve this in Java?

I have tried like this:

Double d = 1.15;
String str = d.toString();
int len = str.substring(str.indexOf(".")).length() - 1;
int i= (int) (d * (long)Math.pow(10,len) % (long)Math.pow(10,len));

But I didn't get the proper answer because when I convert d.toString() the answer is 14.999999999999986.

Answer 1

Try this:

String numberD = String.valueOf(d);
numberD = numberD.substring(numberD.indexOf("."));

Now this numberD variable will have value of 15

Answer 2

Try Math.floor();

double d = 4.24;
System.out.println( d - Math.floor( d )); 

To prevent rounding errors you could convert them to BigDecimal

    double d = 4.24;
    BigDecimal bd = new BigDecimal( d - Math.floor( d ));
    bd = bd.setScale(4,RoundingMode.HALF_DOWN);
    System.out.println( bd.toString() );

Prints 0.2400

Note that the 4 in setScale is the number of digits after the decimal separator ('.')

To have the remainder as an integer value you could modify this to

BigDecimal bd = new BigDecimal(( d - Math.floor( d )) * 100 );
bd = bd.setScale(4,RoundingMode.HALF_DOWN);
System.out.println( bd.intValue() );

Prints 24

Answer 3

Try out RegEx, here ^\d*\., will search for digits followed by dot and replace with blank.

for(double d : new double[]{1.15,0.009,222.9,3.67434}){
    System.out.println(String.valueOf(d).replaceAll("^\\d*\\.",""));
}

gives:

15
009
9
67434

Answer 4

The number after the decimal point is not a well defined concept. For example for "4.24", it could be "0.24", "0.239999999999" (or similar), "24" or "239999999999".

You need to be clear if you are talking about a number or a string of decimal digits ... and whether the input value is a representation of a binary floating point number (in which case "4.24" is most likely an approximation) a decimal floating point number.

Depending on your interpretation, the correct answer will be different.

---

But i didn't get the proper answer because when I converted d.toString() the answer is 14.999999999999986.

You are running up against the problem that double and float are base-2 floating point formats, and in most cases they CANNOT represent decimal floating point numbers precisely. There is no fix for this ... apart from using something like BigDecimal to represent your numbers. (And even then, some loss of precision is possible whenever you do a division or modulo operation.)

Answer 5

If your double is:

d = 1.25;

And if you do this:

String number = String.valueOf(d);

number = number.substring(number.indexOf(".")).substring(1);

then number will be 25.

I don't know if this is the best way to do this, but it works.

Answer 6

If you want the decimal places and you want to round the result.

double d = 4.24;
System.out.printf("%0.2f%n", d - (long) d);

prints

0.24

If you want a rounded value, you have to determine what precision you want.

double d = 4.24;
double fract = d - (long) d;
fract = (long) (fract * 1e9 + 0.5) / 1e9; // round to the 9 decimal places.

Answer 7

    double d = 4.24;
    d = (d - Math.floor( d ));
    int e = (int)(Math.round(d*100));
    System.out.println(e);

Yields 24

Answer 8

  Scanner scr= new Scanner (System.in);
  double d= scr.nextDouble();

  BigDecimal bc= new BigDecimal ((d - Math.floor(d))*1e5); 

  bc=bc.setScale(4,RoundingMode.HALF_DOWN); 

  double f = bc.doubleValue();

    while(f%10==0)
    {
      f=f/10;
    }
  System.out.println(f);

Answer 9

Just as in any other language: multiply by 10 and check the int part of remnant when dividing by 10. The same way you would extract digits from an integer.

short[] digits = new short[max_len];
double tmp = d - ((int) d);
for (int i = 0; i < digits.length && tmp != 0; i++)
{
  tmp *= 10;
  digit[i] = (short) tmp;
  tmp -= (int) tmp;
}

But be aware that for double you need to introduce a precision limit - the maximum number of digits to extract (max_len from the code sample above).

Slight update (added rounding at the edge of precision):

double d = 1.15;
        short[] digits = new short[10];
        double tmp = d - ((int) d) + 0.5 * 1e-10;
        for (int i = 0; i < digits.length && tmp != 0; i++)
        {
            tmp *= 10;
            digits[i] = (short) tmp;
            tmp -= (int) tmp;
        }

Answer 10

Have you tried d = d%1.0? That should return the remainder of dividing d by 1, which should just be the decimal portion.

You could also do this to get the value as an int: d = (int)((d%1.0)*100)

Answer 11

Instead of d.toString(), try this:

String str = String.format("%.2f", d);

Answer 12

try this :

double d = 1.1512;

int i = (int) (d*100 - Math.floor(d)*100);

now this will give you : 15

Answer 13

    ExAMPLE :

    Double d = 469.951479;

    String string = d.toString();

    char[] len = string.substring(string.indexOf(".")).toCharArray();

    System.out.println(len);

Answer 14

double myNumber = 1.15;
//numbers After Dot
double n = 2;
double right_of_dot = (myNumber - (int) myNumber)*Math.pow(10,n);

Answer 15

Try this:

var decimalsOnly = (parseFloat(decNum)-parseInt(decNum,10)).toFoxed(2);