问题描述
为什么一个单precision浮点数有7位precision（或双位数15-16 precision）？
Why does a single-precision floating point number have 7 digit precision (or double 15-16 digits precision)?
任何人能请解释一下我们是如何在这基础上分配给浮动的32位到达（符号（32）指数（30-23），分数（22-0））？
Can anyone please explain how we arrive on that based on the 32 bits assigned for float(Sign(32) Exponent(30-23), Fraction (22-0))?
推荐答案
23分的有效数位（22-0）出现在内存格式，但总precision实际上是24位的，因为我们假设有一领先1。这相当于日志10（2 ^ 24）≈7.225 十进制数字。
23 fraction bits (22-0) of the significand appear in the memory format but the total precision is actually 24 bits since we assume there is a leading 1. This is equivalent to log10(2^24) ≈ 7.225 decimal digits.
双精度型$ P $ pcision浮有52位分数，再加上领先1 53.因此，一个双能容纳日志10（2 ^ 53）≈15.955 十进制数字，不太16。
Double-precision float has 52 bits in fraction, plus the leading 1 is 53. Therefore a double can hold log10(2^53) ≈ 15.955 decimal digits, not quite 16.
注：1领先的是不是一个符号位。它实际上是（ - 1）^ *号1.ffffffff * 2 ^（EEEE常数），但我们不必存储领先1的比例。符号位仍然必须保存
Note: The leading 1 is not a sign bit. It is actually (-1)^sign * 1.ffffffff * 2^(eeee-constant) but we need not store the leading 1 in the fraction. The sign bit must still be stored
有不能被重新psented为2的乘方的总和$ P $一些数字，如1/9：
There are some numbers that cannot be represented as a sum of powers of 2, such as 1/9:
>>>> double d = 0.111111111111111;
>>>> System.out.println(d + "\n" + d*10);
0.111111111111111
1.1111111111111098
如果一个财经节目要做到这一点计算一遍又一遍没有自我修正，但最终将差异。
If a financial program were to do this calculation over and over without self-correcting, there would eventually be discrepancies.
>>>> double d = 0.111111111111111;
>>>> double sum = 0;
>>>> for(int i=0; i<1000000000; i++) {sum+=d;}
>>>> System.out.println(sum);
111111108.91914201
在1十亿求和，我们缺少了$ 2
After 1 billion summations, we are missing over $2.
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