Java中的浮点数

Question

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

Answer 1

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!

Answer 2

因为3.2不能完全表示为浮点数而6.5是（提示：6.5 = 13 * 2 ^（ - 1）），以及3.2是double字面但3.2f是float字面的事实。

Answer 3

Because 3.2f is a float value and 3.2 is a double value. Floating point numbers are always slightly inaccurate because binary representation cannot implement them accurately, so comparing them for exact equality is a bad idea. Particularly comparing floats with doubles. Expressions such as 3.2f == 3.2f are usually okay, but even those can fail in some languages, eg if they represent numbers in registers more accurately than in memory.

Answer 4

这有助于理解

Answer 5

"3.2f" is a of type float. "3.2" is of type double.

Answer 6

另外，我认为任何带小数的内容都默认为double，所以当你进行比较时，你必须添加'f'，如if(f2==6.4f) 。