bigdecimal

Consider the simple test class: import java.math.BigDecimal; /** * @author The Elite Gentleman * */ public class Main { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub BigDecimal x = new BigDecimal("1"); BigDecimal y = new BigDecimal("1.00"); System.out.println(x.equals(y)); System.out.println(x.compareTo(y) == 0 ? "true": "false"); } } You can (consciously) say that x is equal to y (not object reference), but when you run the program, the following result shows: false true Question: What's the difference between compareTo() and equals() in

Java代码 import java.math.BigDecimal; /** * 由于Java的简单类型不能够精确的对浮点数进行运算，这个工具类提供精 * 确的浮点数运算，包括加减乘除和四舍五入。 */ public class Arith{ //默认除法运算精度 private static final int DEF_DIV_SCALE = 10 ; //这个类不能实例化 private Arith(){ } /** * 提供精确的加法运算。 * @param v1 被加数 * @param v2 加数 * @return 两个参数的和 */ public static double add( double v1, double v2){ BigDecimal b1 = new BigDecimal(Double.toString(v1)); BigDecimal b2 = new BigDecimal(Double.toString(v2)); return b1.add(b2).doubleValue(); } /** * 提供精确的减法运算。 * @param v1 被减数 * @param v2 减数 * @return 两个参数的差 */ public static double sub( double v1, double v2){ BigDecimal b1 =

问题的提出： 编译运行下面这个程序会看到什么 [java] view plain copy public class test { public static void main(String args[]) { System.out.println(0.05 + 0.01); System.out.println(1.0 - 0.42); System.out.println(4.015 * 100); System.out.println(123.3 / 100); } }; 你没有看错！结果确实是 [java] view plain copy 0.060000000000000005 0.5800000000000001 401.49999999999994 1.2329999999999999 Java中的简单浮点数类型float和double不能够进行运算。不光是Java，在其它很多编程语言中也有这样的问题。在大多数情况下，计算的结果是准确的，但是多试几次（可以做一个循环）就可以试出类似上面的错误。现在终于理解为什么要有BCD码了。 这个问题相当严重，如果你有9.999999999999元，你的计算机是不会认为你可以购买10元的商品的。 在有的编程语言中提供了专门的货币类型来处理这种情况，但是Java没有。现在让我们看看如何解决这个问题。 解决方案

一、BigDecimal的加减乘除： package com.cy.test.math; import java.math.BigDecimal; public class TestBigDecimal { public static void main(String[] args) { //加法 double d1 = 1.234; double d2 = 2.341; System.out.println(d1 + d2); //3.575 System.out.println(d1 * d2); //2.8887940000000003精度出问题了 System.out.println(d1 / d2); //0.5271251601879539 这个是对的 //使用BigDecimal，加法 BigDecimal b1 = new BigDecimal(Double.toString(d1)); BigDecimal b2 = new BigDecimal(Double.toString(d2)); System.out.println(b1.add(b2).doubleValue()); //3.575 //减法 System.out.println(b1.subtract(b2).doubleValue()); //-1.107 //乘法 System.out

计算数字时，可以使用BigDecimal进行运算。 String number = "100"; BigDecimal decimal1 = new BigDecimal(number); String number1 = "10"; BigDecimal decimal2 = new BigDecimal(number1); System.out.println("加法："+number+"+"+number1+"="+decimal1.add(decimal2).toString()); System.out.println("减法："+number+"-"+number1+"="+decimal1.subtract(decimal2).toString()); System.out.println("乘法："+number+"*"+number1+"="+decimal1.multiply(decimal2).toString()); System.out.println("除法："+number+"/"+number1+"="+decimal1.divide(decimal2).toString()); 以上是四个基本的运算，还有平均数等没有写出来。 有一点需要注意，使用 BigDecimal decimal1 = new BigDecimal(number);

I'm trying to round BigDecimal values up, to two decimal places. I'm using BigDecimal rounded = value.round(new MathContext(2, RoundingMode.CEILING)); logger.trace("rounded {} to {}", value, rounded); but it doesn't do what I want consistently: rounded 0.819 to 0.82 rounded 1.092 to 1.1 rounded 1.365 to 1.4 // should be 1.37 rounded 2.730 to 2.8 // should be 2.74 rounded 0.819 to 0.82 I don't care about significant digits, I just want two decimal places. How do I do this with BigDecimal? Or is there another class/library better suited to this? value = value.setScale(2, RoundingMode.CEILING) 来源

In my little project I need to do something like Math.pow(7777.66, 5555.44) only with VERY big numbers. I came across a few solutions: Use double - but the numbers are too big Use BigDecimal.pow but no support for fractional Use the X^(A+B)=X^A*X^B formula (B is the remainder of the second num), but again no support for big X or big A because I still convert to double Use some kind of Taylor series algorithm or something like that - I'm not very good at math so this one is my last option if I don't find any solutions (some libraries or a formula for (A+B)^(C+D)). Anyone knows of a library or