Why do we need to convert the double into a string, before we can convert it into a BigDecimal?

Question

The source for round in apache commons looks like this:

public static double round(double x, int scale, int roundingMethod) {
    try {
        return (new java.math.BigDecimal(Double.toString(x)).setScale(scale, roundingMethod)).doubleValue();
    } catch (NumberFormatException ex) {
        if (Double.isInfinite(x)) {
            return x;
        } else {
            return Double.NaN;
        }
    }
}

I was wondering, when creating the BigDecimal why did they chose to convert the double to a string (using the Double.toString) instead of simply using the double itself?

In other words, what's wrong with this? :

public static double round(double x, int scale, int roundingMethod) {
    try {
        return (new java.math.BigDecimal(x).setScale(scale, roundingMethod)).doubleValue();
    } catch (NumberFormatException ex) {
        if (Double.isInfinite(x)) {
            return x;
        } else {
            return Double.NaN;
        }
    }
}

Answer 1

It's because the result of BigDecimal(double) constructor is unpredictable as mentioned in javadoc.

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

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.

Test Case:

System.out.println(java.math.BigDecimal.valueOf(0.1).toString());
System.out.println(new java.math.BigDecimal(0.1).toString());