如何获得浮点数的小数部分？ [英] How to get the decimal part of a float?

问题描述

  float n = 22.65f我需要提取一个浮点数的小数部分，但是我得到奇怪的结果： ; 
 //我要x = 0.65f，但是... 
 
 x = n％1; // x = 0.6499996 
 
 x = n  -  Math.floor（n）; // x = 0.6499996185302734 
 
 x = n  - （int）n; // x = 0.6499996 
  

为什么会发生这种情况？为什么我得到这些值而不是 0.65 ？

解决方案

float 只有几位数的精确度，所以你应该期望很容易看到一个圆的错误。尝试 double 这有更多的准确性，但仍有舍入误差。你必须舍弃任何回答你有一个理智的输出。

如果这是不希望的，你可以使用BigDecimal没有舍入错误，但有自己的头痛恕我直言。

编辑：你可能会觉得这很有趣。默认的Float.toString（）使用最小舍入，但通常是不够的。

  System.out.println（With no四舍五入）; 
 float n = 22.65f; 
 System.out.println（n =+ new BigDecimal（n））; 
浮动预期= 0.65f; 
 System.out.println（expected =+ new BigDecimal（expected））; 
 
 System.out.println（n％1 =+ new BigDecimal（n％1））; 
 System.out.println（n  -  Math.floor（n）=+ new BigDecimal（n  -  Math.floor（n）））; 
 System.out.println（n  - （int）n =+ new BigDecimal（n  - （int）n））; 
 
 System.out.println（With rounding）; 
 System.out.printf（n %% 1 =％.2f％n，n％1）; 
 System.out.printf（n  -  Math.floor（n）=％.2f％n，n  -  Math.floor（n））; 
 System.out.printf（n  - （int）n =％.2f％n，n  - （int）n）; 
  

打印

  
n = 22.6499996185302734375 
 expected = 0.64999997615814208984375 $ b $％1 = 0.6499996185302734375 
n  -  Math.floor（n）= 0.6499996185302734375 
n  - （int）n = 0.6499996185302734375 
四舍五入
n％1 = 0.65 
n  -  Math.floor（n）= 0.65 
n  - （int）n = 0.65 
  

I need to extract the decimal part of a float number, but I get weird results:

float n = 22.65f;
// I want x = 0.65f, but...

x = n % 1; // x = 0.6499996

x = n - Math.floor(n); // x = 0.6499996185302734

x = n - (int)n; // x = 0.6499996

Why does this happen? Why do I get those values instead of 0.65?

解决方案

float only has a few digit of precision so you should expect to see a round error fairly easily. try double this has more accuracy but still has rounding errors. You have to round any answer you get to have a sane output.

If this is not desireable you can use BigDecimal which does not have rounding errors, but has its own headaches IMHO.

EDIT: You may find this interesting. The default Float.toString() uses minimal rounding, but often its not enough.

System.out.println("With no rounding");
float n = 22.65f;
System.out.println("n= "+new BigDecimal(n));
float expected = 0.65f;
System.out.println("expected= "+new BigDecimal(expected));

System.out.println("n % 1= "+new BigDecimal(n % 1));
System.out.println("n - Math.floor(n) = "+new BigDecimal(n - Math.floor(n)));
System.out.println("n - (int)n= "+new BigDecimal(n - (int)n));

System.out.println("With rounding");
System.out.printf("n %% 1= %.2f%n", n % 1);
System.out.printf("n - Math.floor(n) = %.2f%n", n - Math.floor(n));
System.out.printf("n - (int)n= %.2f%n", n - (int)n);

Prints

With no rounding
n= 22.6499996185302734375
expected= 0.64999997615814208984375
n % 1= 0.6499996185302734375
n - Math.floor(n) = 0.6499996185302734375
n - (int)n= 0.6499996185302734375
With rounding
n % 1= 0.65
n - Math.floor(n) = 0.65
n - (int)n= 0.65

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