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计算机是如何存储小数的

计算机是如何存储小数的

作者: 公子拙 | 来源:发表于2019-10-10 11:38 被阅读0次

    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 2R
    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

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