Problems with Rule 14–6–2

[James Widman]

Hi all,

The second example in 14–6–2 is:

Code:

template  
void f ( T const & t )
{
   t == t;                       // Non-compliant - Calls NS::operator==
                                 // declared after f
   ::operator ==( t, t );        // Compliant - Calls built-in operator==
   ( operator ==  ) ( t, t ); // Compliant - Calls built-in operator==
}

namespace NS
{
   struct A
   {
      operator int32_t ( ) const;
   };
   bool operator== ( A const &, A const & );
}

int main ( )
{
   NS::A a;
   f ( a );
}

One problem is that the expression '::operator==(t,t)' renders the program ill-formed at template definition time (which means that, if your compiler conforms to ISO C++ 2003, it will issue an error at template definition time and refuse to generate code). This is given by ISO C++ in 14.6 temp.res, which contains this bit of normative text:

If a name does not depend on a template-parameter (as defined in 14.6.2), a declaration (or set of declarations) for that name shall be in scope at the point where the name appears in the template definition; the name is bound to the declaration (or declarations) found at that point and this binding is not affected by declarations that are visible at the point of instantiation.

Because of the way 'operator==' is written, we know that it is not a dependent name. So unqualified name lookup searches for 'operator==' and finds nothing because there is no declaration of that operator before f (either in this example or in the previous example). So the program is ill-formed (diagnostic required).

Likewise, 'operator==' would also render the program ill-formed (if it were not ill-formed already) because there is no operator== template in scope, so the '