一、概述

正整数直接按照源码存储，负整数按照补码存储，那么小数如何存储表示?

二、浮点数的概念(IEEE754)

浮点数是计算机中用于表示实数的一种方法,类似于科学计数法，如:

314.4=3.144x10²
0.00314=3.14x10^-3
255=2.55x10²
由于电脑存储2进制长度有限,这种方法仅能近似的表示某个数。
Java中存放浮点数遵守的标准是IEEE754。

三、IEEE754标准

其定义如何用2进制来存储浮点数,根据公式:
V=(-1)^N x M x 2^R
其中:

N=为符号位,N=1，则为负数;N=0，则为整数
M=底数(尾数)
R=幂(指数)
三部分(NMR)分开存储,有两种方案。

(一)、单精度(32位)

? 1(符号位)+8(指数)+23(底数)，N-(R+127)-M
![][1]

(二)、双精度(64位)

1(符号位)+11(指数)+52(底数)，N-(R+1023)-M
![][2]

(三)、特殊值

无穷 : 指数全为1,底数全为0

0 11111111 000000000000....0000
1 11111111 000000000000....0000

NaN(not a number) : 指数全为1,底数不全为0

0 1111111 101010010...0000
1 1111111 101010000...0000

四、如何存储(如3.14)

(一)、把小数转换为2进制

3.14=0b11.0010001111010111...

(二)、转换为公式格式,底数部分小数点前为1

V=(-1)^N x M x 2^R
0b11.0010001111010111...
=(-1)⁰ x 1.10010001111010111... x 2¹

(三)、符号位N计算

正数为0,负数为1
3.14为正,N=0

(四)、指数部分R计算

R=1
指数部分R要加上127,转换为2进制(无符号)
指数1 加上127(1+127)=128
=0b1000 0000
R=0b1000 0000

(五)、底数部分M计算

去掉小数点前面的1,把剩余的序列拼接即可(如何位数超过范围,会进行四舍五入)
1.10010001111010111...=(丢掉1,这样可以多表示一位数)
M=10010001111010111...

(六)、6)组合起来

3.14=N-R-M
3.14=0b0_10000000_10010001111010111000011

五、总结

浮点数不能准确表示小数
单精度最大值约是3.40E38,最小值(1.40E-45)
双精度最大值约是1.79E308,最小值(4.94E-324)
浮点数如何存储了解即可
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