substitution for std::char_traits<char> in mangled name issue

[zibi at ca.ibm.com]

Hi all,

I just came across an issue with the name mangling schema used be g++ v.
3.2.  Consider the following source:

namespace std {

  template < class C > struct char_traits { };
  template<typename _CharT, typename _Traits> class    basic_ios { };

  template<typename _CharT, typename _Traits = char_traits<_CharT> >
    class basic_ostream : virtual public basic_ios<_CharT, _Traits>
    {
    public:

      typedef basic_ios<_CharT, _Traits>                __ios_type;
      typedef basic_ostream<_CharT, _Traits>            __ostream_type;

      virtual
      ~basic_ostream() { }
      __ostream_type&  operator<<(__ios_type& (*__pf)(__ios_type&)) {
return *((__ostream_type*)0); }

    };
  }

using namespace std;

basic_ios<char, char_traits<char> >& foo(basic_ios<char, char_traits<char>
>& b) { return b; }

int main() {
  basic_ostream<char, char_traits<char> > bo;
  bo << &foo;
  return 0;
}

The mangle name genrated by g++ for std::basic_ostream::operator<< (the one
in bold above) is _ZNSolsEPFRSt9basic_iosIcSt11char_traitsIcEES3_E.  Should
it be
_ZNSolsEPFRSt9basic_iosIcS1_ES4_E?

The reason I question this substitution schema in the name mangling is
because when I  changed the namespace std to stdd then I got :
_ZN4stdd13basic_ostreamIcNS_11char_traitsIcEEElsEPFRNS_9basic_iosIcS2_EES6_E
which indicates that stdd::char_traits<char> has been substituted with S2_
but why std::char_traits<char> is not substituted in the case with std
namespace?  Is this a nonconformance to API spec or there is a special
encoding for std::char_traits<char>?

Regards, ______________________________________
Zbigniew Sarbinowski (Zibi)  C++ for AS/400 developer
Tel.  905-413-6071;  Internet: zibi at ca.ibm.com
8200 Warden Ave. Markham ON, L6G 1C7;  C2/712/8200/MKM