[Bug c++/17805] New: too liberal operator lookup

["rth at gcc dot gnu dot org" <gcc-bugzilla at gcc dot gnu dot org>]

struct  QCString
{
    QCString( int size ) {};
    QCString( const char *str ) {};
};
 
inline bool operator==( const QCString &s1, const QCString &s2 )
{ return 0; }
 
enum EStyle
{
    FS_SOLID
};
 
static EStyle   getFillStyle() { return FS_SOLID; }
 
int main()
{
    if(getFillStyle() == "FS_SOLID") {}
}

As I understand it, this is invalid code.  Alex Oliva writes:

Since one of the operands of == is a class or enumerated type [13.3.1.2]/1,
user-defined operators are candidate functions.  However, to avoid the very kind
of error you've observed, paragraph 3 of the same clause, that defines how the
candidate functions are selected, says, in the second bullet:
   [...] if no operand has a class type, only those non-
    member functions in the lookup set that have a  first  parameter  of
    type  T1 or "reference to (possibly cv-qualified) T1", when T1 is an
    enumeration type, or (if there is a right operand) a second  parame-
    ter of type T2 or "reference to (possibly cv-qualified) T2", when T2
    is an enumeration type, are candidate functions.

Affects at least gcc 3.2 and forward.

-- 
           Summary: too liberal operator lookup
           Product: gcc
           Version: 4.0.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: rth at gcc dot gnu dot org
                CC: gcc-bugs at gcc dot gnu dot org


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=17805