truncate double value to 6 decimal places

Question

I was solving a problem which requires the output to be truncated and to my surprise I haven't been able to find a way to truncate numbers in java.

The output needs to be the a number followed by 6 decimal places.

What I want is double truncate(double number,int places) and the output to be truncate(14/3.0) = 4.666666.

but what I'm getting is

public static double round(double value, int places) {
    if (places < 0) throw new IllegalArgumentException();

    long factor = (long) Math.pow(10, places);
    value = value * factor;
    long tmp = Math.round(value);
    return (double) tmp / factor;
  }
// rounds( 14/3.0 , 6 ) = 4.666667

With String.format I get

String.format("%.6f", 14/3.0) = 4.666667

I also tried a solution that I found on stackoverflow that suggested using BigDecimal and that gave me the very same answer.

NumberFormat also seems to work in the same way

java.text.NumberFormat f = java.text.NumberFormat.getNumberInstance();
f.setMinimumFractionDigits(6);
System.out.println(f.format(14/3.0)); // 4.666667

Is there a built in method that does this or do I need to write my own using something like BigInteger? Is there something I'm doing wrong?

Answer 1

It's because you are not truncating but rounding:

long tmp = Math.round(value);

If you use a cast instead it should work:

long tmp = (long) value;

So the code below outputs 4.666666 as you expect:

public static void main(String[] args) throws Exception {
    System.out.println(truncate(14/3.0, 6));
}

public static double truncate(double value, int places) {
    if (places < 0) {
        throw new IllegalArgumentException();
    }

    long factor = (long) Math.pow(10, places);
    value = value * factor;
    long tmp = (long) value;
    return (double) tmp / factor;
}