https://gcc.gnu.org/bugzilla/show_bug.cgi?id=107813
--- Comment #4 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to Grzegorz Drzewiecki from comment #3)
> (In reply to Jonathan Wakely from comment #1)
> > When you call this->o << t then there is no exact match, so an implicit
> > conversion to unsigned char happens.
>
> OK. But what botter me. I've added global operator like:
>
> std::ostream &operator<<(std::ostream &os, unsigned char c) {
Adding this is undefined behaviour, you're not allowed to redefine what it
means to write fundamental types to std::istream.
> return os << static_cast<unsigned int>(c);
> }
>
> With this one my example works fine:
>
> f << cat << " vs " << dog << "\n\r";
>
> stdout: 67 vs 68.
>
> There is different conversion for either global operator or template
> operator.
No there isn't. You've just changed what the "this->o << t" expression inside
the template operator does. That template operator is still called, but now it
uses a different function to print the enum. You can easily verify this by
stepping through the code in a debugger.
There is no GCC bug here, this is just how C++ works.
