java - 为什么我们需要先将 double 转换为字符串，然后才能将其转换为 BigDecimal？

Question

round 的来源在 apache commons 中看起来像这样:

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;
        }
    }
}

我想知道，在创建 BigDecimal 时，为什么他们选择将 double 转换为字符串(使用 Double.toString)而不是简单地使用 double 本身？

换句话说，这有什么问题？ :

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

这是因为 BigDecimal(double) 构造函数的结果如 javadoc 中所述是不可预测的。

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.

测试用例:

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