Hello! I'm implementing boost::lexical_cast for class A. When operator<< is declared in the same namespace, as class A, it works: * #include <boost/lexical_cast.hpp> * #include <string> * * namespace abc { * * struct A { * int a; * }; * * std::ostream& operator<<(std::ostream& o, const A& a) { * return o << a.a; * } * * } * * int main() { * abc::A a; * std::string a_str = boost::lexical_cast<std::string>(a); * } But if operator<< is declared in root namespace, it will not compile: * #include <boost/lexical_cast.hpp> * #include <string> * * namespace abc { * * struct A { * int a; * }; * * } * * std::ostream& operator<<(std::ostream& o, const abc::A& a) { * return o << a.a; * } * * int main() { * abc::A a; * std::string a_str = boost::lexical_cast<std::string>(a); * } Part of error message: In file included from error.cpp:2:0: /usr/include/boost/lexical_cast.hpp: In member function ‘bool boost::detail::lexical_stream<Target, Source, Traits>::operator<<(const Source&) [with Target = std::basic_string<char>, Source = abc::A, Traits = std::char_traits<char>]’: /usr/include/boost/lexical_cast.hpp:1151:13: instantiated from ‘Target boost::detail::lexical_cast(typename boost::call_traits<Source>::param_type, CharT*, size_t) [with Target = std::basic_string<char>, Source = abc::A, bool Unlimited = true, CharT = char, typename boost::call_traits<Source>::param_type = const abc::A&, size_t = long unsigned int]’ /usr/include/boost/lexical_cast.hpp:1174:77: instantiated from ‘Target boost::lexical_cast(const Source&) [with Target = std::basic_string<char>, Source = abc::A]’ error.cpp:19:59: instantiated from here /usr/include/boost/lexical_cast.hpp:595:48: error: no match for ‘operator<<’ in ‘((boost::detail::lexical_stream<std::basic_string<char>, abc::A, std::char_traits<char>
*)this)->boost::detail::lexical_stream<std::basic_string<char>, abc::A, std::char_traits<char> >::stream << input’ /usr/include/c++/4.5/ostream:108:7: note: candidates are:
So, it does not see my declaration. However, <code>std::cout << a << std::endl;</code> works in both cases. What is the reason of this error? Why operator<< can not be declared in root namespace? Sorry for my question. Maybe, it is connected less to boost than to c++. Regards.