Java中的浮点数 [英] Float numbers in Java

问题描述

  public static void main（ String [] args）
 {
 
 float f1 = 3.2f; 
 float f2 = 6.5f; 
 
 if（f1 == 3.2）
 System.out.println（same）; 
 else 
 System.out.println（different）; 
 
 if（f2 == 6.5）
 System.out.println（same）; 
 else 
 System.out.println（different）; 
 
 code $ 
 
 $ b $ o不同
相同
 
解决方案
 6.5有一个有限的二进制表达式：110.1 
 
 
位可以完美地表示这个数字。 
 
 110.100000000000000000000（float）
 
 = 6.5 
 $ b $ 110.10000000000000000000000000000000000000000000000000（双精度）
 
 = 6.5 
 
 3.2另一方面有一个无限的二进制表示：101.0011001100110011 ... 
 
 
 float和double没有无限精度，因此只能接近这个数字：（
 
 101.001100110011001100110（float）
 
 = 3.2000000476837158203125 
 > 
 
 101.00110011001100110011001100110011001100110011001101（double）
 
 = 3.20000000000000017763568394002504646778106689453125 
 
 
正如您可以清楚地看到的，这些数字不是一样！
Could anyone please me why the output of the following programme is not " different different"?
public static void main(String[] args)
{

float f1=3.2f;
float f2=6.5f;

if(f1==3.2)
System.out.println("same");
else
System.out.println("different");

if(f2==6.5)
System.out.println("same");
else
System.out.println("different");
}
o/p :different
same
解决方案
6.5 has a finite binary representation: 110.1

Any floating type with at least 4 significant bits can represent this number perfectly. 

110.100000000000000000000 (float)

= 6.5

110.10000000000000000000000000000000000000000000000000 (double)

= 6.5

3.2 on the other hand has an infinite binary representation: 101.0011001100110011...

float and double don't have infinite precision and thus can only approximate this number :(

101.001100110011001100110 (float)

= 3.2000000476837158203125

101.00110011001100110011001100110011001100110011001101 (double)

= 3.20000000000000017763568394002504646778106689453125

As you can clearly see, these numbers are not the same!

                        
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