EOS 智能合约 plublic key 转换

 

 

在做一个EOS 的action接口时,定义如下:

void setbplist(const account_name bp_name, const uint64_t bp_time, const std::string hash_bplist, const std::vector< uosio::producer_key >& bplist);

   其中有一个结构uosio::producer_key,它在合约的结构中是如下定义:

struct /*  */producer_key {
      account_name     producer_name;
      public_key       block_signing_key;

      friend bool operator < ( const producer_key& a, const producer_key& b ) {
         return a.producer_name < b.producer_name;
      }

      UOSLIB_SERIALIZE( producer_key, (producer_name)(block_signing_key))
   };

  其中 public_key 定义如下:

  struct public_key {
      unsigned_int        type;
      std::array<char,33> data;

      friend bool operator == ( const public_key& a, const public_key& b ) {
        return std::tie(a.type,a.data) == std::tie(b.type,b.data);
      }
      friend bool operator != ( const public_key& a, const public_key& b ) {
        return std::tie(a.type,a.data) != std::tie(b.type,b.data);
      }
      UOSLIB_SERIALIZE( public_key, (type)(data) )
   };

  这里的 public_key 是一个压缩成33个字节的数据,但其实我传的是 一组公匙,我想在智能合约里还原成公匙的形态。EOS公匙是基于base58编码,这个与比特币的一致,这里不再详述了。

 

  下面下面是我的代码:

std::string system_contract::get_public_key_from_array(const char* pub, const int  len) {
   std::string result; 
   
   checksum160 ck;
   ripemd160(pub, len, &ck);

   std::vector<unsigned char> vec;

   for(int i = 0; i < len; ++i) {
      vec.push_back(pub[i]);
   }

   for(int i = 0; i < 4; ++i ) {
      vec.push_back(ck.hash[i]);
   }

   result = "UOS" + EncodeBase58(vec);

   print("result: ", result, "\n");

   return result;
}

std::string system_contract::EncodeBase58(const std::vector<unsigned char>& vch)
{
  return EncodeBase58(vch.data(), vch.data() + vch.size());
}

std::string system_contract::EncodeBase58(const unsigned char* pbegin, const unsigned char* pend)
{
  // Skip & count leading zeroes.
  int zeroes = 0;
  int length = 0;
  while (pbegin != pend && *pbegin == 0) {    
    pbegin++;
    zeroes++;
  }
  // Allocate enough space in big-endian base58 representation.
  int size = (pend - pbegin) * 138 / 100 + 1; // log(256) / log(58), rounded up.
  std::vector<unsigned char> b58(size);
  // Process the bytes.
  while (pbegin != pend) {
    int carry = *pbegin;
    int i = 0;
    // Apply "b58 = b58 * 256 + ch".
    for (std::vector<unsigned char>::reverse_iterator it = b58.rbegin(); (carry != 0 || i < length) && (it != b58.rend()); it++, i++) {
      carry += 256 * (*it);
      *it = carry % 58;
      carry /= 58;
    }

    assert(carry == 0);
    length = i;
    pbegin++;
  }


  // Skip leading zeroes in base58 result.
  std::vector<unsigned char>::iterator it = b58.begin() + (size - length);
  while (it != b58.end() && *it == 0)
    it++;


  // Translate the result into a string.
  std::string str;
  str.reserve(zeroes + (b58.end() - it));
  str.assign(zeroes, '1');


  while (it != b58.end())
    str += pszBase58[*(it++)];

  return str;
}

  

 

我的合约里调用此接口的部分代码:

 

接口的调用:

 cluos -u http://10.186.11.112:8000 push action uosio setbplist '{"bp_name":"marsaccount3","bp_time":"1553685636", "hash_bplist":"9d3329c56e7e6f6068f7dbd905f4b99ffa86ba53897c2a399f5d3637ade41c9a", "bplist":[{"producer_name":"marsaccount3", "block_signing_key":"UOS6aWfdf6tHWCepZpP6MzdynuNMAkNKr6nbNMguTuCatq88LyG4G"},{"producer_name":"dragonexsafe", "block_signing_key":"UOS7DVNg9bsq1zUZtUbWwA1UyhaqFBmNrCRVnPFeGYgsx7kzfAtiC"},{"producer_name":"uosvegasjack", "block_signing_key":"UOS5aTdkbRaPH5WKYPJaxX8HfqGtX8hKx8p6FDPAzpkyJiYnuBu5c"}]}' -x 2000  -p marsaccount3@active

 

可以看到我的三个公匙是:

UOS6aWfdf6tHWCepZpP6MzdynuNMAkNKr6nbNMguTuCatq88LyG4G
UOS7DVNg9bsq1zUZtUbWwA1UyhaqFBmNrCRVnPFeGYgsx7kzfAtiC
UOS5aTdkbRaPH5WKYPJaxX8HfqGtX8hKx8p6FDPAzpkyJiYnuBu5c

但实际到了EOS合约里面,已经被处理过了,像第一个公匙 UOS6aWfdf6tHWCepZpP6MzdynuNMAkNKr6nbNMguTuCatq88LyG4G 在传到合约里对应的就是33个字节的数据,以十六进制打印出来就是  02de9288d61afea6084227b2db9449fd58986652bcda91a4efeeac3ef0ee4ad80c

通过以上的接口就还原公匙的原型了。

 

参考来源:https://www.liaoxuefeng.com/article/001523444489789210f7acb0355485d91cfb1a44788ac9b000

posted @ 2019-03-27 19:48  HBright  阅读(515)  评论(0编辑  收藏  举报