![7999999999999998]( "7999999999999998")
本文介绍了双倍减去双精度后的精度损失的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
好吧我有以下代码块：
int rotation = e.getWheelRotation（）;
if（rotation zoom（zoom + rotation * -.05）;
else if（zoom - .05> 0）
zoom（zoom - rotation * .05）;
System.out.println（zoom）;
现在，缩放变量是double类型的，初始设置为1。结果是1 - .05 = .95; .95 - .05 = .9; .9 - .05 = .85;等等。这似乎并不是这样，虽然当我打印结果，你可以看到下面：
- 0.95
- 0.8999999999999999
- 0.8499999999999999
- 0.7999999999999998
- 0.7499999999999998
- 0.6999999999999997
希望有人能够清楚地解释。我搜索了互联网，我读到它有一些限制，当我们以二进制存储浮动，但我还是不太明白。我的问题的解决方案并不是很重要，但是我想理解这种行为。 解决方案
Java使用浮点数。他们不是很精确。着名的例子是：
System.out.println（0.1d + 0.2d）;
...输出 0.30000000000000004 / p>
您所看到的只是这种不精确的症状。你可以通过使用 double 而不是 float 来提高精确度。
如果你正在处理财务计算，你可能更喜欢 to float 或 double 。
Alright so I've got the following chunk of code:
int rotation = e.getWheelRotation();
if(rotation < 0)
zoom(zoom + rotation * -.05);
else if(zoom - .05 > 0)
zoom(zoom - rotation * .05);
System.out.println(zoom);
Now, the zoom variable is of type double, initially set to 1. So, I would expect the results to be like 1 - .05 = .95; .95 - .05 = .9; .9 - .05 = .85; etc. This appears to be not the case though when I print the result as you can see below:
- 0.95
- 0.8999999999999999
- 0.8499999999999999
- 0.7999999999999998
- 0.7499999999999998
- 0.6999999999999997
Hopefully someone is able to clearly explain. I searched the internet and I read it has something to do with some limitations when we're storing floats in binary but I still don't quite understand. A solution to my problem is not shockingly important but I would like to understand this kind of behavior.
解决方案
Java uses IEEE-754 floating point numbers. They're not perfectly precise. The famous example is:
System.out.println(0.1d + 0.2d);
...which outputs 0.30000000000000004.
What you're seeing is just a symptom of that imprecision. You can improve the precision by using double rather than float.
If you're dealing with financial calculations, you might prefer BigDecimal to float or double.
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