分割和乘法交替过程中的Java精度 [英] Java precision during division and multiplication alterneting process

问题描述

可能重复：

如何解决Java Rounding双重问题

请帮助，
我编程一些Java中的计算器。我使用双重类型。双数位在小数点后15位。我有以下问题：

Please Help, I programm some calculator in Java. I use double type. Double has 15 digits after the decimal point. I have problem with the following:

1/3 * 3 = 0.9999999999999999

1/3 * 3 = 0.9999999999999999

我需要1/3 * 3 = 1

I need 1/3 * 3 = 1

我如何解决这个问题？
我保持结果为Double。与其他数学运算相同的问题，例如

How can I solve this problem? I keep result in Double. The same problem I have with other mathematical operations, for example

sqrt（6）= 2.449489742783，接下来我平方的结果，我得到：5.999999999999999

sqrt(6) = 2.449489742783, and next I square the result and I get: 5.999999999999999

推荐答案

您正在处理浮点运算的固有局限性。

You're dealing with inherent limitations of floating-point arithmetic.

• 阅读文章每个计算机科学家应该知道的知识浮点数算术。


• 对于平等检查，您应该使用类似于 abs（xy）< epsilon 而不是 x == y


• 为了显示，你应该舍入到最接近的小数位你真的在乎。

• Read the paper What Every Computer Scientist Should Know About Floating-Point Arithmetic.
• For equality-checking, you should be using something like abs(x-y) < epsilon rather than x == y
• For display purposes, you should round to the nearest decimal place that you actually care about.

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