将编码的std :: string从Base16转换为Base10？ [英] Convert encoded std::string from Base16 to Base10?

问题描述

我有一个以Base16编码的大整数的 std :: string ：
$ b

  bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63 
 e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287 
 f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4 
 5af06cd8554fd55f1c442c2ed 
  

我想要一个以Base10编码的大整数的 std :: string ：

  13182363340585954094154991955162141609757130565683854218475776626603716062690 
 50741824486137510938646762753180989129520441058729412931959771922633699694948 
 46611764803267065720664398942078304585998290003537553345030144535441671492050 
 01138054588415687622649540474976282005406232907125282540703919964112809484362 
 9分配
  

如何从Base16转换字符串t Base10？

解决方案

将Base16编码的std :: string转换为Base10？

以下内容适用于您。下面的代码向您展示了如何使用C风格的字符串，这很容易概念化。您在将CryptoPP :: Integer转换为LPCTSTR 的前一个问题有引用。

  #include< iostream> 
 #include< string> 
使用namespace std; 
 
 #includecryptlib.h
 #includeinteger.h
 using namespace CryptoPP; 
 
 INT主（INT的argc，字符* argv的[]）
 {
串S2，S1 = 
 的bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63 
 的e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287 
f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4
5af06cd8554fd55f1c442c2ed; 
 
 //追加'h'表示Base16 
 // Integer n（（s1 +h）。c_str（））; 
 
 //预先加上'0x'来表示Base16 
整数n（（0x+ s1）.c_str（））; 
 
 //转换为Base10 
 s2 = IntToString< Integer>（n，10）; 
 cout<< s2<< ENDL; 
 
返回0; 
 
 
 
 
 
 $ b $ p $上面的代码显示了你如何用C风格的字符串做到这一点，这很容易概念化。另一种方法是使用Crypto ++   Pipeline  将ASCII字符串转换为大端字节数组。 
  #include< iostream> 
 #include< string> 
使用namespace std; 
 
 #includecryptlib.h
 #includeinteger.h
 #includefilters.h
 #includehex.h
使用名称空间CryptoPP; 
 
 INT主（INT的argc，字符* argv的[]）
 {
串S3，S2，S1 = 
 的bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63 
 e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287 
f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4
5af06cd8554fd55f1c442c2ed; 
 
 //使用HexDecoder转换为big-endian数组
 StringSource ss（s1，true，new HexDecoder（new StringSink（s2）））; 
 
 //使用big-endian数组构造n 
整数n（（const byte *）s2.data（），s2.size（））; 
 
 //转换为Base10 
 s3 = IntToString<整数>（n，10）; 
 cout<< s3<< ENDL; 
 
返回0; 
 
  
 
 
 
 
 
 
这是另一种执行方式使用Crypto ++   Pipeline  进行转换。 
  #include< iostream> 
 #include< string> 
使用namespace std; 
 
 #includecryptlib.h
 #includeinteger.h
 #includefilters.h
 #includehex.h
使用名称空间CryptoPP; 
 
 INT主（INT的argc，字符* argv的[]）
 {
串S2，S1 = 
 的bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63 
 的e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287 
f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4
5af06cd8554fd55f1c442c2ed; 
 
 //使用源代码转换为big-endian数组
 StringSource ss（s1，true，new HexDecoder）; 
 
 //使用big-endian数组构造n 
整数n; 
 n.Decode（ss，ss.MaxRetrievable（））; 
 
 //转换为Base10 
 s2 = IntToString< Integer>（n，10）; 
 cout<< s2<< ENDL; 
 
返回0; 
 
  
 
 
 
 
 
 
如果您对将ASCII字符串转换为内部表示的字节数组的算法，然后参见  StringToInteger 在 integer.cpp  中。它反复按基数（2,8,10,16等）分。 
I have a std::string with a large integer encoded in Base16:
bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63
e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287
f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4
5af06cd8554fd55f1c442c2ed
I want a std::string with a large integer encoded in Base10:
13182363340585954094154991955162141609757130565683854218475776626603716062690
50741824486137510938646762753180989129520441058729412931959771922633699694948
46611764803267065720664398942078304585998290003537553345030144535441671492050
01138054588415687622649540474976282005406232907125282540703919964112809484362
9
How do I convert the strings from Base16 to Base10?
解决方案

  
Convert encoded std::string from Base16 to Base10?
The following should work for you. The code below shows you how to do it with C-style strings, which is easy to conceptualize. Your previous question at Convert CryptoPP::Integer to LPCTSTR has the references.
#include <iostream>
#include <string>
using namespace std;

#include "cryptlib.h"
#include "integer.h"
using namespace CryptoPP;

int main(int argc, char* argv[])
{
  string s2, s1 =
      "bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63"
      "e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287"
      "f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4"
      "5af06cd8554fd55f1c442c2ed";

  // Append 'h' to indicate Base16
  // Integer n((s1 + "h").c_str());

  // Prepend '0x' to indicate Base16
  Integer n(("0x" + s1).c_str());

  // Convert to Base10
  s2 = IntToString<Integer>(n, 10);
  cout << s2 << endl;

  return 0;
}




The code above shows you how to do it with C-style strings, which is easy to conceptualize. Another way to do it uses a Crypto++ Pipeline to convert the ASCII string into a big-endian array of bytes. 
#include <iostream>
#include <string>
using namespace std;

#include "cryptlib.h"
#include "integer.h"
#include "filters.h"
#include "hex.h"
using namespace CryptoPP;

int main(int argc, char* argv[])
{
  string s3, s2, s1 =
      "bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63"
      "e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287"
      "f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4"
      "5af06cd8554fd55f1c442c2ed";

  // Use a HexDecoder to convert to big-endian array
  StringSource ss(s1, true, new HexDecoder(new StringSink(s2)));

  // Use big-endian array to construct n
  Integer n((const byte*)s2.data(), s2.size());

  // Convert to Base10
  s3 = IntToString<Integer>(n, 10);
  cout << s3 << endl;

  return 0;
}




Here's another way to perform the conversion using a Crypto++ Pipeline.
#include <iostream>
#include <string>
using namespace std;

#include "cryptlib.h"
#include "integer.h"
#include "filters.h"
#include "hex.h"
using namespace CryptoPP;

int main(int argc, char* argv[])
{
  string s2, s1 =
      "bbb91c1c95b656f386b19ab284b9c0f66598e7761cd71569734bb72b6a7153b77613a6cef8e63"
      "e9bd9bb1e0e53a0fd8fa2162b160fcb7b461689afddf098bfc32300cf6808960127f1d9f0e287"
      "f948257f7e0574b56585dd1efe1192d784b9c93f9c2215bd4867062ea30f034265374fa013ab4"
      "5af06cd8554fd55f1c442c2ed";

  // Use a source to convert to big-endian array
  StringSource ss(s1, true, new HexDecoder);

  // Use big-endian array to construct n
  Integer n;
  n.Decode(ss, ss.MaxRetrievable());

  // Convert to Base10
  s2 = IntToString<Integer>(n, 10);
  cout << s2 << endl;

  return 0;
}




If you are interested in the algorithm that converts the ASCII string to a byte array for internal representation, then see StringToInteger in integer.cpp. It repeatedly divides by the base (2, 8, 10, 16, etc).

                        
这篇关于将编码的std :: string从Base16转换为Base10？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！