c++ - 使用 stringstreams 将字符串转换为 __uint128_t

Question

我正在尝试从字符串中提取不同类型的数据。

void                    readHeader(char buf[BUFFSIZE])
{
  std::istringstream    hdr(buf);
  __uint128_t           id_client;

  hdr >> id_client; // doesn't compile
}

当我执行 hdr >> id_client 时出现此错误:

Unix/UnixSocket.cpp:158:10: error: ambiguous overload for ‘operator>>’ in ‘hdr >> id_client’ Unix/UnixSocket.cpp:158:10: note: candidates are: In file included from /usr/include/c++/4.7/sstream:39:0,
                 from Unix/UnixSocket.cpp:11: /usr/include/c++/4.7/istream:118:7: note: std::basic_istream<_CharT,
_Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__istream_type& (*)(std::basic_istream<_CharT, _Traits>::__istream_type&)) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_istream<_CharT,
_Traits>::__istream_type = std::basic_istream<char>] <near match> /usr/include/c++/4.7/istream:118:7: note:   no known conversion for argument 1 from ‘__int128 unsigned’ to ‘std::basic_istream<char>::__istream_type& (*)(std::basic_istream<char>::__istream_type&) {aka std::basic_istream<char>& (*)(std::basic_istream<char>&)}’ /usr/include/c++/4.7/istream:122:7: note: std::basic_istream<_CharT,
_Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__ios_type& (*)(std::basic_istream<_CharT, _Traits>::__ios_type&)) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_istream<_CharT,
_Traits>::__istream_type = std::basic_istream<char>; std::basic_istream<_CharT, _Traits>::__ios_type = std::basic_ios<char>] <near match> /usr/include/c++/4.7/istream:122:7:

有什么方法可以将我的 id_client 正确存储在这个 __uint128_t 变量中吗？

Answer 1

https://gmplib.org/可能有帮助。 gmpxx 对象抽象的 mpz_class 类支持 I/O 运算符，而 mpz_export(...) 函数允许您将结果转换为字节数组。如果它们超过 16 个字节，您可能会抛出异常或以其他方式提示。不是很快，但我想实现起来很快。