Use of operator T const& to get a reference to self

Given the following program:

    #include <locale>

    class FG
    {
    public:
        explicit FG( std::locale const& l )
                : myLoc( &l ) {}

        template< typename F >
                operator F const&() const
        {
            return get< F >( *myLoc, &std::use_facet ) ;
        }

    private:
        std::locale const* myLoc ;

        template< typename F >
        F const& get( std::locale const& l,
                      F const& (*f)( std::locale const& ) ) const
        {
            return (*f)( l ) ;
        }
    } ;

    FG
    getF( std::locale const& l )
    {
        return FG( l ) ;
    }

    void
    f()
    {
        std::ctype< char > const& ct = getF( std::locale() ) ;
    }

G++ (4.1.0) instantates the template operator F const& to get
the const reference needed to call the copy constructor (which
of course fails to compile, since use_facet<FG> is not legal).
Providing a non-template:

    operator FG const&() const { return *this ; }

solves the problem, but is g++ correct here? I would have
expected the temporary "FG( l )" to bind directly to the const
reference of the (compiler generated) copy constructor.

Or maybe my question is: is this intentional? Given ?8.5.3/5,
"[...]If the initializer expression [...]-- has a class type [it
does] and can be implicitly converted to an lvalue of type "cv3
T3", where "cv1 T1" is reference-compatible with "cv3 T3" [...]
then [...] the reference is bound to the lvalue result of the
conversion [...]" But it doesn't seem at all natural to have a
user defined conversion called for a copy.

--
James Kanze (GABI Software) mailto:james.kanze@gmail.com
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