问题描述

我想知道在 Java 中修复精度错误的最佳方法是什么.正如您在以下示例中看到的那样，存在精度错误:

I'm wondering what the best way to fix precision errors is in Java. As you can see in the following example, there are precision errors:

class FloatTest
{
  public static void main(String[] args)
  {
    Float number1 = 1.89f;
    
    for(int i = 11; i < 800; i*=2)
    {
      System.out.println("loop value: " + i);
      System.out.println(i*number1);
      System.out.println("");
    }
  }
}

显示的结果是:

循环值:11

20.789999

循环值:22

41.579998

循环值:44

83.159996

循环值:88

166.31999

循环值:176

332.63998

循环值:352

665.27997

循环值:704

1330.5599

另外，如果有人能解释为什么它只从 11 开始，并且每次都将价值翻倍.我认为所有其他值(或至少其中许多值)都显示了正确的结果.

Also, if someone can explain why it only does it starting at 11 and doubling the value every time. I think all other values (or many of them at least) displayed the correct result.

这样的问题过去让我很头疼，我通常使用数字格式化程序或将它们放入字符串中.

Problems like this have caused me headache in the past and I usually use number formatters or put them into a String.

正如人们所提到的，我可以使用双精度，但在尝试之后，似乎 1.89 作为双倍乘以 792 仍然会输出错误(输出为 1496.8799999999999).

As people have mentioned, I could use a double, but after trying it, it seems that 1.89 as a double times 792 still outputs an error (the output is 1496.8799999999999).

我想我会尝试其他解决方案，例如 BigDecimal

I guess I'll try the other solutions such as BigDecimal

推荐答案

如果你真的很在意精度，你应该使用 BigDecimal

If you really care about precision, you should use BigDecimal

https://docs.oracle.com/javase/8/docs/api/java/math/BigDecimal.html

https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/math/BigDecimal.html

这篇关于Java中浮点数的精度错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！