原始双精度值是否等于幅值? [英] Primitive double value equal depends on magnitude?

问题描述

我必须检查两个双精度值是否相等，包括大小和精度.我遇到了一个奇怪的情况，即原始的double equals校验不一致，并且取决于值的大小.

I have to check to see whether two double values are equal including magnitude and precision. I encounter a weird scenario where primitive double equals check is not consistent and depend on magnitude of value.

我使用过的Java版本:

Java version i've used:

java version "1.6.0_26"
Java(TM) SE Runtime Environment (build 1.6.0_26-b03)
Java HotSpot(TM) Client VM (build 20.1-b02, mixed mode, sharing)

我的代码:

public class Test{
     public static void main(String args[]) throws Exception{
         String val1 = "15.999999999999999";
         String val2 = "16";
         String val3 = "16.999999999999999";
         String val4 = "17";
         double d1 = Double.parseDouble(val1);
         double d2 = Double.parseDouble(val2);
         double d3 = Double.parseDouble(val3);
         double d4 = Double.parseDouble(val4);
         System.out.println(val1 + "=" + val2 + "? ===>" + (d1==d2));
         System.out.println(val3 + "=" + val4 + "? ===>" + (d3==d4));
     }
}

输出:

15.999999999999999=16? ===>false
16.999999999999999=17? ===>true

推荐答案

由于各种无聊而又复杂的原因，您尝试执行的操作将无法正常工作，您可以在此处阅读以下内容:

What you're trying to do won't work for various boring and complicated reasons that you can read about here: http://download.oracle.com/docs/cd/E19957-01/806-3568/ncg_goldberg.html

除非您要比较双精度的位表示(可能比较有见或没有见识)，否则始终需要某种epsilon值，即处理数字浮点表示时的误差余量.像这样:

Unless you're going to compare the bit-representation of the doubles (which may or may not be insightful), you will always need some sort of epsilon value, i.e. margin of error when dealing with floating-point representations of numbers. Something like:

boolean doublesAreEqual(double d1, double d2)
{
    double d = d1 / d2;
    return (Math.abs(d - 1.0) < 0.00001 /* epsilon */);
}

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