问题描述

我需要处理一个二进制数.

I need to work with a binary number.

我试着写:

const x = 00010000;

但是没有用.

我知道我可以使用与 00010000 具有相同值的十六进制数，但我想知道 C++ 中是否有二进制数的类型，如果没有，是我的问题还有其他解决方案吗?

I know that I can use an hexadecimal number that has the same value as 00010000, but I want to know if there is a type in C++ for binary numbers and if there isn't, is there another solution for my problem?

推荐答案

您可以在等待 C++0x 时使用 BOOST_BINARY.:) BOOST_BINARY 可以说比模板实现有优势，因为它也可以在 C 程序中使用(它是 100% 预处理器驱动的.)

You can use BOOST_BINARY while waiting for C++0x. :) BOOST_BINARY arguably has an advantage over template implementation insofar as it can be used in C programs as well (it is 100% preprocessor-driven.)

要进行相反的操作(即以二进制形式打印出一个数字)，您可以使用不可移植的 itoa 函数，或 实现您自己的.

To do the converse (i.e. print out a number in binary form), you can use the non-portable itoa function, or implement your own.

不幸的是，您不能使用 STL 流进行 base 2 格式化(因为 setbase 仅支持 8、10 和 16 基数)，但您可以使用 itoa 的 std::string 版本，或者(更简洁但效率稍低)std::bitset.

Unfortunately you cannot do base 2 formatting with STL streams (since setbase will only honour bases 8, 10 and 16), but you can use either a std::string version of itoa, or (the more concise, yet marginally less efficient) std::bitset.

#include <boost/utility/binary.hpp>
#include <stdio.h>
#include <stdlib.h>
#include <bitset>
#include <iostream>
#include <iomanip>

using namespace std;

int main() {
  unsigned short b = BOOST_BINARY( 10010 );
  char buf[sizeof(b)*8+1];
  printf("hex: %04x, dec: %u, oct: %06o, bin: %16s
", b, b, b, itoa(b, buf, 2));
  cout << setfill('0') <<
    "hex: " << hex << setw(4) << b << ", " <<
    "dec: " << dec << b << ", " <<
    "oct: " << oct << setw(6) << b << ", " <<
    "bin: " << bitset< 16 >(b) << endl;
  return 0;
}

产生:

hex: 0012, dec: 18, oct: 000022, bin:            10010
hex: 0012, dec: 18, oct: 000022, bin: 0000000000010010

另请阅读 Herb Sutter 的庄园农场的字符串格式化程序 进行有趣的讨论.

Also read Herb Sutter's The String Formatters of Manor Farm for an interesting discussion.

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