通过隐式转换为字符串流对象时，过载解析失败

Question

考虑以下代码：#include <string>#include <ostream>#include <iostream>struct NameType {   operator std::string() { return "wobble"; }};struct Person {   NameType name;};int main() {   std::cout << std::string("bobble");   std::cout << "wibble";   Person p;   std::cout << p.name;}它产生于GCC 4.3.4以下：prog.cpp: In function ‘int main()’:prog.cpp:18: error: no match for ‘operator<<’ in ‘std::cout << p.Person::name’/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:112: note: candidates are: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>& (*)(std::basic_ostream<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>]/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:121: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ios<_CharT, _Traits>& (*)(std::basic_ios<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>]/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:131: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::ios_base& (*)(std::ios_base&)) [with _CharT = char, _Traits = std::char_traits<char>]/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:169: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long int) [with _CharT = char, _Traits = std::char_traits<char>]/usr/lib/gcc/i686-pc-linux-gnu/4.3.4/include/g++-v4/ostream:173: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long unsigned int) [with _CharT = char, _Traits = std::char_traits<char>]op<<(ostream&, string const&)为何免费不将其纳入过载组？这是由于所需的重载是模板实例化和... ADL的结合吗？

Answer 1

在C ++ 98中为14.8.1 / 4

如果参数类型不包含参与模板参数推导的模板参数，则对函数参数执行隐式转换（第4节），以将其转换为相应的函数参数的类型。

在这里您想实例化

template <class charT, class traits, class Allocator>

  basic_ostream<charT, traits>&

    operator<<(basic_ostream<charT, traits>&,

               const basic_string<charT, traits, Allocator>&);

在没有明确提供任何模板参数的情况下进行推断。因此，所有参数都包含一个模板参数，该参数参与模板参数推导，因此它们都无法从隐式转换中获取其值。