Re: [Boost-users] [lexical_cast][variant] Compile failed when included boost.variant.

Subject: Re: [Boost-users] [lexical_cast][variant] Compile failed when included boost.variant.
From: Vitaly Budovski (vbudovski+news_at_[hidden])
Date: 2012-01-31 00:00:04

On 31 January 2012 14:43, Tianjiao Yin <ytj000_at_[hidden]> wrote:
> Hello, all:
>
> I met a strange compile error, this code will reproduce the problem::
>
> Â  Â #include <boost/lexical_cast.hpp>
> Â  Â #include <boost/variant.hpp>
> Â  Â #include <iostream>
>
> Â  Â namespace foo {
> Â  Â struct bar {};
> Â  Â }
>
> Â  Â std::ostream& operator << (std::ostream& out, foo::bar) {
> Â  Â  Â  Â return out << "BAR";
> Â  Â }
>
> Â  Â int main() {
> Â  Â  Â  Â boost::lexical_cast<std::string>(foo::bar());
> Â  Â }
>
> However, if I just remove that unused include boost.variant, it works::
>
> Â  Â #include <boost/lexical_cast.hpp>
> Â  Â //#include <boost/variant.hpp>
> Â  Â #include <iostream>
>
> Â  Â namespace foo {
> Â  Â struct bar {};
> Â  Â }
>
> Â  Â std::ostream& operator << (std::ostream& out, foo::bar) {
> Â  Â  Â  Â return out << "BAR";
> Â  Â }
>
> Â  Â int main() {
> Â  Â  Â  Â boost::lexical_cast<std::string>(foo::bar());
> Â  Â }
>
> Another solution is put that "operator<<" in namespace foo::
>
> Â  Â #include <boost/lexical_cast.hpp>
> Â  Â #include <boost/variant.hpp>
> Â  Â #include <iostream>
>
> Â  Â namespace foo {
> Â  Â struct bar {};
> Â  Â std::ostream& operator << (std::ostream& out, foo::bar) {
> Â  Â  Â  Â return out << "BAR";
> Â  Â }
> Â  Â } // namespace foo
>
> Â  Â int main() {
> Â  Â  Â  Â boost::lexical_cast<std::string>(foo::bar());
> Â  Â }
>
> 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
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> Boost-users_at_[hidden]
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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.