On 31 January 2012 14:43, Tianjiao Yin <ytj000@gmail.com> wrote:

Hello, all:

I met a strange compile error, this code will reproduce the problem::

   #include <boost/lexical_cast.hpp>    #include <boost/variant.hpp>    #include <iostream>

   namespace foo {    struct bar {};    }

   std::ostream& operator << (std::ostream& out, foo::bar) {        return out << "BAR";    }

   int main() {        boost::lexical_cast<std::string>(foo::bar());    }

However, if I just remove that unused include boost.variant, it works::

   #include <boost/lexical_cast.hpp>    //#include <boost/variant.hpp>    #include <iostream>

   namespace foo {    struct bar {};    }

   std::ostream& operator << (std::ostream& out, foo::bar) {        return out << "BAR";    }

   int main() {        boost::lexical_cast<std::string>(foo::bar());    }

Another solution is put that "operator<<" in namespace foo::

   #include <boost/lexical_cast.hpp>    #include <boost/variant.hpp>    #include <iostream>

   namespace foo {    struct bar {};    std::ostream& operator << (std::ostream& out, foo::bar) {        return out << "BAR";    }    } // namespace foo

   int main() {        boost::lexical_cast<std::string>(foo::bar());    }

Anyway, what is boost.variant's business here? Won't that fail to compile, or am I missing something? BTW, I'm using boost 1.48.0 and gcc 4.6.2 in Linux 3.2.2.

Regards, Tianjiao _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users

Your overloaded operator isn't found by ADL, which looks for it in namespaces std and foo. You should put it in the same namespace as your struct.