有没有人知道任何常用的C ++库，它提供了编码和解码从基数到基数32的数字的方法，反之亦然？
谢谢， 斯特凡诺
答案 0 :(得分：4)
[已更新] 显然，C ++ std::setbase() IO操纵器和普通<<和>> IO操作员只处理基数8 ，10和16，因此对于处理基座32是无用的。
所以要解决你的转换问题
你需要求助于其他功能。
为了将包含基本2..36表示的C样式字符串转换为整数，您可以使用#include <cstdlib>并使用strtol(3)＆amp;公司功能集。
至于将整数转换为具有任意基数的字符串......我找不到简单的答案。 printf(3)样式格式字符串仅处理基础8,10,16 AFAICS，就像std::setbase一样。任何人吗？
答案 1 :(得分：2)
您的意思是“基数为10到32”，而不是整数到base32 吗？后者似乎更有可能也更有用;默认情况下，标准格式化的I / O函数在处理整数时会生成基本的10字符串格式。
对于基数32到整数转换，标准库strtol（）函数将执行此操作。对于互惠，你不需要一个你自己可以轻松实现的东西的库（不是所有东西都是乐高积木）。
这是一个例子，不一定是最有效的，但很简单;
#include <cstring>
#include <string>
long b32tol( std::string b32 )
{
return strtol( b32.c_str(), 0, 32 ) ;
}
std::string itob32( long i )
{
unsigned long u = *(reinterpret_cast<unsigned long*>)( &i ) ;
std::string b32 ;
do
{
int d = u % 32 ;
if( d < 10 )
{
b32.insert( 0, 1, '0' + d ) ;
}
else
{
b32.insert( 0, 1, 'a' + d - 10 ) ;
}
u /= 32 ;
} while( u > 0 );
return b32 ;
}
#include <iostream>
int main()
{
long i = 32*32*11 + 32*20 + 5 ; // BK5 in base 32
std::string b32 = itob32( i ) ;
long ii = b32tol( b32 ) ;
std::cout << i << std::endl ; // Original
std::cout << b32 << std::endl ; // Converted to b32
std::cout << ii << std::endl ; // Converted back
return 0 ;
}
答案 2 :(得分：2)
直接回答原始（现在是旧的）问题，我不知道在base32中编码字节数组的任何公共库，或者之后再次解码它们。但是，我上周提出需要将base32中表示的SHA1哈希值解码为原始字节数组。这里有一些C ++代码（带有一些值得注意的Windows /小端文件），我写的就是这样做，并验证结果。
请注意，与上面的Clifford代码相比，如果我没有弄错，假设RFC 4648中提到了“base32hex”字母，我的代码假定为“base32”字母（“AZ”和“2-7”） ）。
// This program illustrates how SHA1 hash values in base32 encoded form can be decoded
// and then re-encoded in base16.
#include "stdafx.h"
#include <string>
#include <vector>
#include <iostream>
#include <cassert>
using namespace std;
unsigned char Base16EncodeNibble( unsigned char value )
{
if( value >= 0 && value <= 9 )
return value + 48;
else if( value >= 10 && value <= 15 )
return (value-10) + 65;
else //assert(false);
{
cout << "Error: trying to convert value: " << value << endl;
}
return 42; // sentinal for error condition
}
void Base32DecodeBase16Encode(const string & input, string & output)
{
// Here's the base32 decoding:
// The "Base 32 Encoding" section of http://tools.ietf.org/html/rfc4648#page-8
// shows that every 8 bytes of base32 encoded data must be translated back into 5 bytes
// of original data during a decoding process. The following code does this.
int input_len = input.length();
assert( input_len == 32 );
const char * input_str = input.c_str();
int output_len = (input_len*5)/8;
assert( output_len == 20 );
// Because input strings are assumed to be SHA1 hash values in base32, it is also assumed
// that they will be 32 characters (and bytes in this case) in length, and so the output
// string should be 20 bytes in length.
unsigned char *output_str = new unsigned char[output_len];
char curr_char, temp_char;
long long temp_buffer = 0; //formerly: __int64 temp_buffer = 0;
for( int i=0; i<input_len; i++ )
{
curr_char = input_str[i];
if( curr_char >= 'A' && curr_char <= 'Z' )
temp_char = curr_char - 'A';
if( curr_char >= '2' && curr_char <= '7' )
temp_char = curr_char - '2' + 26;
if( temp_buffer )
temp_buffer <<= 5; //temp_buffer = (temp_buffer << 5);
temp_buffer |= temp_char;
// if 8 encoded characters have been decoded into the temp location,
// then copy them to the appropriate section of the final decoded location
if( (i>0) && !((i+1) % 8) )
{
unsigned char * source = reinterpret_cast<unsigned char*>(&temp_buffer);
//strncpy(output_str+(5*(((i+1)/8)-1)), source, 5);
int start_index = 5*(((i+1)/8)-1);
int copy_index = 4;
for( int x=start_index; x<(start_index+5); x++, copy_index-- )
output_str[x] = source[copy_index];
temp_buffer = 0;
// I could be mistaken, but I'm guessing that the necessity of copying
// in "reverse" order results from temp_buffer's little endian byte order.
}
}
// Here's the base16 encoding (for human-readable output and the chosen validation tests):
// The "Base 16 Encoding" section of http://tools.ietf.org/html/rfc4648#page-10
// shows that every byte original data must be encoded as two characters from the
// base16 alphabet - one charactor for the original byte's high nibble, and one for
// its low nibble.
unsigned char out_temp, chr_temp;
for( int y=0; y<output_len; y++ )
{
out_temp = Base16EncodeNibble( output_str[y] >> 4 ); //encode the high nibble
output.append( 1, static_cast<char>(out_temp) );
out_temp = Base16EncodeNibble( output_str[y] & 0xF ); //encode the low nibble
output.append( 1, static_cast<char>(out_temp) );
}
delete [] output_str;
}
int _tmain(int argc, _TCHAR* argv[])
{
//string input = "J3WEDSJDRMJHE2FUHERUR6YWLGE3USRH";
vector<string> input_b32_strings, output_b16_strings, expected_b16_strings;
input_b32_strings.push_back("J3WEDSJDRMJHE2FUHERUR6YWLGE3USRH");
expected_b16_strings.push_back("4EEC41C9238B127268B4392348FB165989BA4A27");
input_b32_strings.push_back("2HPUCIVW2EVBANIWCXOIQZX6N5NDIUSX");
expected_b16_strings.push_back("D1DF4122B6D12A10351615DC8866FE6F5A345257");
input_b32_strings.push_back("U4BDNCBAQFCPVDBL4FBG3AANGWVESI5J");
expected_b16_strings.push_back("A7023688208144FA8C2BE1426D800D35AA4923A9");
// Use the base conversion tool at http://darkfader.net/toolbox/convert/
// to verify that the above base32/base16 pairs are equivalent.
int num_input_strs = input_b32_strings.size();
for(int i=0; i<num_input_strs; i++)
{
string temp;
Base32DecodeBase16Encode(input_b32_strings[i], temp);
output_b16_strings.push_back(temp);
}
for(int j=0; j<num_input_strs; j++)
{
cout << input_b32_strings[j] << endl;
cout << output_b16_strings[j] << endl;
cout << expected_b16_strings[j] << endl;
if( output_b16_strings[j] != expected_b16_strings[j] )
{
cout << "Error in conversion for string " << j << endl;
}
}
return 0;
}
答案 3 :(得分：1)
我不知道任何常用的base32编码库，但Crypto ++包含public domain base32 encoder and decoder。