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如何在C / C ++中获得大数字的精确二进制表示形式？

[英]How to get exact binary representation of big numbers in c/c++?

原文 2017-11-29 14:38:50 0 1 c/ binary/ long-integer/ biginteger/ scientific-notation

I am trying to convert big numbers to binary in C/C++ and Java but if i take input in plain decimal like 998446744073709551615 output is correct but if i will use scientific notation ex : 1.7334e+32 then binary representation comes wrong. 我正在尝试在C / C ++和Java中将大数转换为二进制，但如果我采用普通的十进制形式输入，例如998446744073709551615输出是正确的，但是如果我将使用科学记数法ex： 1.7334e+32则二进制表示形式会出错。

I have tested from double and BigDecimals from binary in C/C++ and Java. 我已经在C / C ++和Java中从double和BigDecimals从二进制进行了测试。

String to Long Double in C/C++ Test : https://ideone.com/EeOyNP 在C / C ++测试中将字符串转换为Long Double： https : //ideone.com/EeOyNP

String to Big Decimal in Java Test : https://ideone.com/OAvx7q 在Java测试中将字符串转换为大十进制： https : //ideone.com/OAvx7q

The problem is with numbers which are represented in more than 64 bits aren't represented in scientific notation somehow . 问题在于用64 bits以上的数字表示的数字没有以科学的记号表示。

Check out the output below. 查看下面的输出。

Output from C/C++ Code : C / C ++代码的输出：

Input = 998446744073709551615 Expected Binary = 1101100010000000111011011001111011110011001010110011111111111111111111 输入= 998446744073709551615预期二进制= 1101100010000000111011011001111011110011001010110011111111111111111111111

Output : Successfully parsed strtold (C-Style): 9.98447e+20 Binary : 1101100010000000111011011001111011110011001010110100000000000000000000 输出：成功解析strtold（C样式）：9.98447e + 20二进制：1101100010000000111011011001111011110011001010110100000000000000000000

stringstream parsed stringstream (C++ Style): 9.98447e+20 Binary : 1101100010000000111011011001111011110011001010110100000000000000000000 stringstream解析的stringstream（C ++样式）：9.98447e + 20 Binary：1101100010000000111011011001111011110011001010110100000000000000000000

showBitDiff statistics : Total Bits 70 Bits 49 Bits matched 21 Bits not matched showBitDiff统计信息：总位70位49位匹配21位不匹配

Output from Java Code : Java代码的输出：

Decimal String Part : 小数字符串部分：

decimalString : 998446744073709551615 Scientific notation : 9.984467440737096E20 decimalString：998446744073709551615科学计数法：9.984467440737096E20

Decimal-String Radix Info : 十进制字符串基数信息：
Binary : 1101100010000000111011011001111011110011001010110011111111111111111111 Decimal : 998446744073709551615 Hexa : 0x36203B67BCCACFFFFF Bit length : 70 二进制：1101100010000000111011011001111011110011001010110011111111111111111111十进制：998446744073709551615十六进制：0x36203B67BCCACFFFFF位长度：70

Exponent String Part : 指数字符串部分：
exponentString : 9.984467440737096E20 exponentString：9.984467440737096E20

Exponent String Radix Info : 指数字符串基数信息：
Binary : 1101100010000000111011011001111011110011001010110100001011110100000000 Decimal : 998446744073709600000 Hexa : 0x36203B67BCCAD0BD00 Bit length : 70 二进制：1101100010000000111011011001111011110011001010110100001011110100000000十进制：998446744073709600000十六进制：0x36203B67BCCAD0BD00位长度：70

Both BigInts are not equal 两个BigInts不相等

How do i solve this problem and represent big numbers correctly in C/C++? 如何解决此问题并在C / C ++中正确表示大数字？ i don't want solution in java i just used java for testing purpose because it has bigDecimal Class for very large arbitrary numbers thanks. 我不想要Java解决方案，我只是将Java用于测试目的，因为它具有针对非常大的任意数字的bigDecimal类，谢谢。

1 个解决方案

If you first convert to double to represent your decimal in exponent representation then of course it is wrong. 如果您首先转换为double以指数形式表示小数，那当然是错误的。 Have a look at https://en.wikipedia.org/wiki/Floating-point_arithmetic . 看看https://en.wikipedia.org/wiki/Floating-point_arithmetic 。 TLDR; TLDR； If you have more different numbers than states (2^128) you will have gaps. 如果数字比州（2 ^ 128）多得多，您将有空白。 Thats what double is designed for from the beginning. 那就是从一开始就为double设计的。 If you require an exact representations of your actual number do not convert it into a floating point representation. 如果您需要实际数字的精确表示，请不要将其转换为浮点表示。