一、概述

正整數(shù)直接按照源碼存儲，負(fù)整數(shù)按照補(bǔ)碼存儲，那么小數(shù)如何存儲表示?

二、浮點(diǎn)數(shù)的概念(IEEE754)

浮點(diǎn)數(shù)是計(jì)算機(jī)中用于表示實(shí)數(shù)的一種方法,類似于科學(xué)計(jì)數(shù)法，如:

• 314.4=3.144x10²
• 0.00314=3.14x10^-3
• 255=2.55x10²
  由于電腦存儲2進(jìn)制長度有限,這種方法僅能近似的表示某個(gè)數(shù)。
  Java中存放浮點(diǎn)數(shù)遵守的標(biāo)準(zhǔn)是IEEE754。

三、IEEE754標(biāo)準(zhǔn)

其定義如何用2進(jìn)制來存儲浮點(diǎn)數(shù),根據(jù)公式:
V=(-1)^N x M x 2^R
其中:

• N=為符號位,N=1，則為負(fù)數(shù);N=0，則為整數(shù)
• M=底數(shù)(尾數(shù))
• R=冪(指數(shù))
  三部分(NMR)分開存儲,有兩種方案。

(一)、單精度(32位)

? 1(符號位)+8(指數(shù))+23(底數(shù))，N-(R+127)-M
![][1]

(二)、雙精度(64位)

1(符號位)+11(指數(shù))+52(底數(shù))，N-(R+1023)-M
![][2]

(三)、特殊值

• 無窮 : 指數(shù)全為1,底數(shù)全為0

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

• NaN(not a number) : 指數(shù)全為1,底數(shù)不全為0

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

四、如何存儲(如3.14)

(一)、把小數(shù)轉(zhuǎn)換為2進(jìn)制

3.14=0b11.0010001111010111...

(二)、轉(zhuǎn)換為公式格式,底數(shù)部分小數(shù)點(diǎn)前為1

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

(三)、符號位N計(jì)算

正數(shù)為0,負(fù)數(shù)為1
3.14為正,N=0

(四)、指數(shù)部分R計(jì)算

R=1
指數(shù)部分R要加上127,轉(zhuǎn)換為2進(jìn)制(無符號)
指數(shù)1 加上127(1+127)=128
=0b1000 0000
R=0b1000 0000

(五)、底數(shù)部分M計(jì)算

去掉小數(shù)點(diǎn)前面的1,把剩余的序列拼接即可(如何位數(shù)超過范圍,會進(jìn)行四舍五入)
1.10010001111010111...=(丟掉1,這樣可以多表示一位數(shù))
M=10010001111010111...

(六)、6)組合起來

3.14=N-R-M
3.14=0b0_10000000_10010001111010111000011

五、總結(jié)

1. 浮點(diǎn)數(shù)不能準(zhǔn)確表示小數(shù)
2. 單精度最大值約是3.40E38,最小值(1.40E-45)
3. 雙精度最大值約是1.79E308,最小值(4.94E-324)
4. 浮點(diǎn)數(shù)如何存儲了解即可
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