Round a double in Java
floating-point, java
Solution
You have to replace
BigDecimal bigDecimal = new BigDecimal(d);
with
BigDecimal bigDecimal = BigDecimal.valueOf(d);
and you will get the expected results:
2.66
1.66
Explanation from Java API:
BigDecimal.valueOf(double val) - uses the double's canonical string representation provided by the Double.toString() method. This is preferred way to convert a double (or float) into a BigDecimal.
new BigDecimal(double val) - uses the exact decimal representation of the double's binary floating-point value and thus results of this constructor can be somewhat unpredictable.
Problem
I have found this great solution for rounding: ``` static Double round(Double d, int precise) { BigDecimal bigDecimal = new BigDecimal(d); bigDecimal = bigDecimal.setScale(precise, RoundingMode.HALF_UP); return bigDecimal.doubleValue(); } ``` However, the results are confusing: ``` System.out.println(round(2.655d,2)); // -> 2.65 System.out.println(round(1.655d,2)); // -> 1.66 ``` Why is it giving this output? I'm using jre 1.7.0_45.