Java Double get all Digits after dot/comma

Question

it´sa simple task but i´m not able to solve it on my own..

i got

    double digit1 = 12.1;
    double digit2 = 12.99;

and need a method which gives me this:

    anyMethod(digit1); //returns 10
    anyMethod(digit2); //returns 99

what i have is

public static void getAfterComma(double digit) {

    BigDecimal bd = new BigDecimal(( digit - Math.floor( digit )) * 100 );
    bd = bd.setScale(4,RoundingMode.HALF_DOWN);
    System.out.println(bd.toBigInteger()); // prints digit1=1 and digit2=99

} 

anyway i prefer integer as the returntype.. anybody got a quick solution/tip?

kindly

Answer 1

Why not you simply use:

int anyMethod(double a){
  //if the number has two digits after the decimal point.
  return (int)((a + 0.001) * 100) % 100;
}

Answer 2

A simple way of getting the fractional part of a double is to use the modulo operator, % . However, you'll have to take into account the fact that floating point arithmetic isn't precise. For example,

System.out.println(12.1 % 1);   // outputs 0.09999999999999964
System.out.println(12.99 % 1);  // outputs 0.9900000000000002

If you want to get two decimal digits as an int , which is what I think you're asking, you can achieve this, glossing over the floating point issues, like so:

System.out.println(Math.round((12.1 % 1) * 100));   // outputs 10
System.out.println(Math.round((12.99 % 1) * 100));  // outputs 99

However, you should consider going further down the BigDecimal path you started down, which uses arbitrary precision arithmetic. You could do something like this:

System.out.println(new BigDecimal("12.1").remainder(BigDecimal.ONE));   // outputs 0.1
System.out.println(new BigDecimal("12.99").remainder(BigDecimal.ONE));   // outputs 0.99

If, as before, you want two decimal digits from this, you can do this:

System.out.println(new BigDecimal("12.1").remainder(BigDecimal.ONE).multiply(new BigDecimal(100)).setScale(2, RoundingMode.HALF_UP).intValue());   // outputs 0.1
System.out.println(new BigDecimal("12.99").remainder(BigDecimal.ONE).multiply(new BigDecimal(100)).setScale(2, RoundingMode.HALF_UP).intValue());   // outputs 0.99

Note that there a couple of differences between these last two methods and the first two: they preserve the sign of the argument, so if you use the final example for -12.99, you'll get -99 back, and they treat the fractional part of an integer as 1, so if you use the final example for 12, you'll get 100 back.

Answer 3

It's not necessary to use Number tyeps all the time. You can take advantage of String as a mediator.

String mediator = Double.valueOf(d1).toString();
mediator = mediator.substring(mediator.indexOf('.') + 1);
Integer result = Integer.valueOf(mediator);

Answer 4

Try this out

 public static void main(String args[]){
        double a=12.99;
        double b=12.1;


        System.out.println(method(a));
        System.out.println(method(b));
    }

    private static int method(double a) {
        return (int) ((a*100)%100);

    }    

Answer 5

If I understand correctly, you need to return n digits after the dot for a given double number. So... let's see:

public int decimalDigits(double x, int n) {
    double ans;
    ans = (x - (int) x) * Math.pow(10, n);
    return (int) ans;
}

Done. Hope this helps you.

---

For your specific example, 'n = 2' should do.

Answer 6

sachin-pasalkar done it! little fix but fine!

public static int anyMethod(double a){
      return (int) (a*100)%100;
    }

Answer 7

check out this code returns digits after '.' always. Without any extra parameters other than double variable.

public int anyMethod(double d)
{
    String numString = d+"";
    return Integer.parseInt(numString.substring(numString.indexOf('.')+1));
}