Question

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

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;
}

这适用于我尝试的所有类型，除了char的运算符＆lt;＆lt;被认为是含糊不清的。我无法理解为什么它适用于int，short或任何其他具有运算符定义的类型，但不适用于chars。有人有什么想法吗？

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 ==========

Answer 1

当重载决议出现时，编译器似乎将const char&到char的转换放在与向上转换std::ostringstream到std::ostream相同的级别上。

解决方案可能是模拟operator<<的类型以避免向上转换：

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

std :: ostream＆amp;的模糊过载operator＆lt;＆lt;（std :: ostream＆amp; sstr，const T＆amp; val）

1 个答案: