Lambda表达式mapToDouble.sum精度问题

Posted 2022-01-04

public static void main(String[] args) List<Double> doubles = Arrays.asList(new Double(3.3), new Double(3.3), new Double(3.3)); double sum = doubles.stream().mapToDouble(Double::doubleValue).sum(); System.out.println(sum);// 9.899999999999999 System.out.println(formatDigit(sum,2));// 9.9 System.out.println(formatDigit_down(sum, 2)); // 9.89 List<Double> doubles1 = Arrays.asList(new Double(1.1), new Double(1.1), new Double(1.1)); double sum1 = doubles1.stream().mapToDouble(Double::doubleValue).sum(); System.out.println(sum1);// 3.3000000000000003 System.out.println(formatDigit(sum1,2));// 3.3 System.out.println(formatDigit_down(sum1, 2));// 3.3 /** * 将数据转换为保留指定小数位数（0，1，2）格式的数，四舍五入 * */ public static double formatDigit(double num, int decimalPlace) DecimalFormat fm = null; switch (decimalPlace) case 0: fm = new DecimalFormat("##"); break; case 1: fm = new DecimalFormat("##.#"); break; case 2: fm = new DecimalFormat("##.##"); break; case 3: fm = new DecimalFormat("##.###"); break; case 4: fm = new DecimalFormat("##.####"); break; default: break; if (fm == null) return num; StringBuffer sbf = new StringBuffer(); fm.format(num, sbf, new FieldPosition(java.text.NumberFormat.FRACTION_FIELD)); return Double.parseDouble(sbf.toString()); /** * 将数据转换为保留指定小数位数（0，1，2）格式的数。向下取值：如 5.567 -> 5.56 -5.567 -> -5.56 */ public static double formatDigit_down(double num, int decimalPlace) DecimalFormat fm = null; switch (decimalPlace) case 0: fm = new DecimalFormat("##"); break; case 1: fm = new DecimalFormat("##.#"); break; case 2: fm = new DecimalFormat("##.##"); break; case 3: fm = new DecimalFormat("##.###"); break; case 4: fm = new DecimalFormat("##.####"); break; default: break; if (fm == null) return num; StringBuffer sbf = new StringBuffer(); fm.setRoundingMode(RoundingMode.DOWN);// 向下取值设置、 fm.format(num, sbf, new FieldPosition(java.text.NumberFormat.FRACTION_FIELD)); return Double.parseDouble(sbf.toString());