std::ostream&operator< & lt; (std:: ostream&压力，压力& &;val)

Ambiguous overload of std::ostream& operator<<(std::ostream& sstr, const T& val)

本文关键字：压力 ostream std val lt operator      更新时间：2023-10-16

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，该函数利用来自单独命名空间的重载std::ostream& operator<<;

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型，short型或任何其他定义了操作符的类型，但不适用于char。有人有什么想法吗?

1>application_srcgeneral_experiments.cpp(39): error C2593: 'operator <<' is ambiguous
1>          application_srcgeneral_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 11.0VCincludeostream(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 11.0VCincludeostream(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 11.0VCincludeostream(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_srcgeneral_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 ==========

当重载解析来临时，编译器似乎将从const char&到char的转换与从std::ostringstream到std::ostream的转换放在同一级别上。

解决方案可以是模板化operator<<的类型，以避免上行转换:

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