Round a double in Java

Question

I have found this great solution for rounding:

static Double round(Double d, int precise) {
    BigDecimal bigDecimal = new BigDecimal(d);
    bigDecimal = b         


        

Answer 1

As said in API

1. The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.

2. The String constructor, on the other hand, is perfectly predictable: writing new BigDecimal("0.1") creates a BigDecimal which is exactly equal to 0.1, as one would expect. Therefore, it is generally recommended that the String constructor be used in preference to this one.

3. When a double must be used as a source for a BigDecimal, note that this constructor provides an exact conversion; it does not give the same result as converting the double to a String using the Double.toString(double) method and then using the BigDecimal(String) constructor. To get that result, use the static valueOf(double) method.

It's because of cannot represent double value exactly. So you have to use BigDecimal bigDecimal = BigDecimal.valueOf(d); instead of BigDecimal bigDecimal = new BigDecimal(d);