2

为了序列化任何对象（即对于没有全局ostream& operator<<的对象，字符串为空），我创建了一个函数，该函数利用了std::ostream& operator<<来自单独命名空间的重载；

namespace _impl {
template <typename T>
std::ostream& operator<<(std::ostream& osstr, const T& val) {
  return osstr;
}
}

template <typename T>
std::string serialize_any(const T& val) {
  using namespace _impl;
  std::ostringstream osstr;
  osstr<< val;
  std::string str(osstr.str());
  return str;
}

这适用于我尝试过的所有类型，除了 char 的 operator<< 被认为是模棱两可的。我不明白为什么它适用于 int's、short's 或任何其他定义了运算符的类型，但不适用于 chars。有人有想法么？

1>application_src\general_experiments.cpp(39): error C2593: 'operator <<' is ambiguous
1>          application_src\general_experiments.cpp(26): could be 'std::ostream &_impl::operator <<<T>(std::ostream &,const T &)'
1>          with
1>          [
1>              T=char
1>          ]
1>          C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\include\ostream(914): or       'std::basic_ostream<_Elem,_Traits> &std::operator <<<char,std::char_traits<char>>(std::basic_ostream<_Elem,_Traits> &,_Elem)' [found using argument-dependent lookup]
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>
1>          ]
1>          C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\include\ostream(827): or       'std::basic_ostream<_Elem,_Traits> &std::operator <<<std::char_traits<char>>(std::basic_ostream<_Elem,_Traits> &,char)' [found using argument-dependent lookup]
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>
1>          ]
1>          C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\include\ostream(742): or       'std::basic_ostream<_Elem,_Traits> &std::operator <<<char,std::char_traits<char>>(std::basic_ostream<_Elem,_Traits> &,char)' [found using argument-dependent lookup]
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>
1>          ]
1>          while trying to match the argument list '(std::ostringstream, const char)'
1>          application_src\general_experiments.cpp(52) : see reference to function template instantiation 'std::string serialize_any<char>(const T &)' being compiled
1>          with
1>          [
1>              T=char
1>          ]
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
4

1 回答 1

1

当重载决议到来时，编译器似乎将转换 from const char&to置于与向上转换tochar相同的级别。std::ostringstreamstd::ostream

解决方案可能是模板类型operator<<以避免向上转换：

namespace _impl {
    template <typename T, typename Y>
    Y& operator<<(Y& osstr, const T& val) {
      return osstr;
    }
}
于 2013-04-15T14:56:08.360 回答