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
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
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