Convert encoded std::string from Base16 to Base10?

问题:

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?

回答1:

Since you don't want a library, you need some code. It's not a trivial problem, but not too complex either. Let's start with my Bignum class from another answer and add a couple of functions to it.

class Bignum
{
    //...

    Bignum& operator+=(int rhs)
    {
        assert(rhs >= 0 && rhs <= 999999999);
        uint32_t carry = rhs;
        for (size_t i = 0; i < parts.size(); i++)
        {
            uint32_t sum = parts[i] + carry;
            parts[i] = (uint32_t)(sum % 1000000000UL);
            carry = (uint32_t)(sum / 1000000000UL);
        }
        if (carry != 0)
            parts.push_back(carry);
        return *this;
    }

    void FromHex(const char * pString)
    {
        while (*pString != 0)
        {
            char ch = toupper(*pString++);
            assert((ch >= '0' && ch <= '9') || (ch >= 'A' && ch <= 'F'));
            int digit = (ch <= '9') ? (ch - '0') : (ch - 'A' + 10);
            *this *= 16;
            *this += digit;
        }
    }

See the whole thing in action: http://coliru.stacked-crooked.com/a/cb5061a00c945875

回答2:

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).