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

From:
"=?iso-8859-1?q?Daniel_Kr=FCgler?=" <daniel.kruegler@googlemail.com>
Newsgroups:
comp.std.c++
Date:
Thu, 29 Mar 2007 13:52:02 CST
Message-ID:
<1175194229.830718.296910@b75g2000hsg.googlegroups.com>
James Kanze schrieb:

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.


An interesting question, but IMO the standard is quite clear
here, because 12.3.2/1 says unambigiously:

"[..] A conversion function is never used to convert a (possibly
cv-qualified) object to the (possibly cv-qualified) same object
type (or a reference to it),[..]"

This sentence exists since the 1998 standard and has also
not changed in the current draft. I considered for a while that
gcc might not view a temporary as an object, but this seems
also rather clearly expressed in both 1.8 and 12.2.

I observed that even my quite old mingw gcc 3.4, a rather old
Borland Compiler (I just can't find its version number) and
MS-VS2003 show the unexpected behaviour you mention.
But no newer compiler I tested (including all public available
Comeau online versions and MS-VS2005-SP1) did. Of course
we have no majoritarian decision here, but I tend to say that
the standard is unusually unmistakable here ;-)

All newer compilers follow rather strict the mentioned rule:
To enforce this, I simply added the operator FG(FG&) to your
class and none (except the older ones) tried the conversion
operator but gave up with something like "class "FG" has no
suitable copy constructor".

As a not very relevant sidenode: Even by using the language
extension mode of MS-VS2005 the compiler did not do that
but selected instead the actually forbidden FG(FG&) c'tor with
a corresponding warning

"nonstandard extension used : 'argument' : conversion from 'FG'
to 'FG &' A non-const reference may only be bound to an lvalue;
copy constructor takes a reference to non-const") warning).

Concluding, I would say that a bug report for gcc seems a
reasonable choice.

Greetings from Bremen,

Daniel

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