使用无理数.返回1/22140.简化?

问题描述

这不可能吗?应该简化分数以避免不终止小数扩展"公共类方法{

Is this not possible? Should the fraction be simplified to avoid 'non terminating decimal expansion'public class method {

BigDecimal a = new BigDecimal("1");
BigDecimal b = new BigDecimal ("22140");
BigDecimal c;


void cal () {

    System.out.println(c=a.divide(b));

}

推荐答案

我假设您正在询问是否可以将1/22140完全表示为 BigDecimal .答案是否定的. BigDecimal 类不能表示具有非终止十进制扩展数的值，并且也没有简化允许它的方程式.

I assume you are asking if it is possible to represent 1 / 22140 exactly as a BigDecimal. The answer to that is No. The BigDecimal class cannot represent values with a non-terminating decimal expansion, and there is no simplification of the equation that will allow it.

问:为什么?

A:基本上是数学.1/22140不能表示为m * 10 ，其中m和s是有限整数.

A: Basically, mathematics. 1 / 22140 cannot be represented as m * 10 where m and s are finite integers.

有两种实用的替代方法:

There are two practical alternative approaches:

1. 进行除法运算时，可以使用 MathContext 自变量来指定如何对数字取整以及保持多少精度；例如

1. You can use a MathContext argument when you do the division to specify how to round the number and how much precision to keep; e.g.

c = a.divide(b, new MathContext(100, RoundingMode.HALF_UP));

这将计算精度为100个十进制数字的结果，并四舍五入(以传统方式)到最接近的可表示值.

This will compute the result with 100 decimal digits of precision, rounding (in the conventional way) to the nearest representable value.

请注意，这种方法为您提供 inexact 结果；例如(1/3)* 3将不完全是1，无论您在 MathContext 中将精度设置为什么.

Note that this approach gives you inexact results; e.g. (1 / 3) * 3 won't be exactly 1, no matter what you set the precision to in the MathContext.

您可以找到并使用第三方的 Rational 或 Fraction 类，或者编写自己的类.

You can find and use a 3rd-party Rational or Fraction class, or write your own.

请注意，Java SE不提供此类.

Note that Java SE doesn't provide such a class.

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