Re: Hidden overloaded operator in multiple inheritance
"Carl Daniel [VC++ MVP]" wrote:
I found in multiple inheritance only the first base
class's overloaded
operators are visible. My overloaded operators has
different
signatures in the different base classes. They are not
ambiguous.
'using operator..' statements can bring them back. Please
see my
minimal example below. I'd like to know if it is a bug or
a feature.
What version of VC++ are you using?
Comeau C++, which is generally considered the best
reference on standards conformance agrees that the calls
in your code are ambiguous. Interestingly,. VC8 (VS 2005)
SP1 compiles your code without error. I'm inclined to
agree with VC8 and yourself that these calls are not
ambiguous, but I haven't really gone through the code and
the overload rules with a fine toothed comb to back that
opinion up with quotes from the standard.
I tried the example code with VC++ 2005 _pre_ SP1 and it
fails exactly as OP pointed. I think that Comeau is right,
though. Here's the relevant part from the Standard:
<quote>
10.2/2 Member name lookup
The following steps define the result of name lookup in a
class scope, C. First, every declaration for the name in the
class and in each of its base class sub-objects is
considered. A member name f in one subobject B hides a
member name f in a sub-object A if A is a base class
sub-object of B. Any declarations that are so hidden are
eliminated from consideration. Each of these declarations
that was introduced by a using-declaration is considered to
be from each sub-object of C that is of the type containing
the declaration designated by the using-declaration.96) If
the resulting set of declarations are not all from
sub-objects of the same type, or the set has a nonstatic
member and includes members from distinct sub-objects, there
is an ambiguity and the program is ill-formed. Otherwise
that set is the result of the lookup.
</quote>
I think that the sentense before last one describes the
problem. Let's take `comp1' as an example:
std::string s = c1 + std::string (" world!");
After name lookup the following member functions are in the
set:
int basei<comp1>::operator + (int i)
std::string bases<comp1>::operator + (std::string s)
Now, the above set contains declaration of nonstatic members
from distinct sub-objects. The same behaviour can be
demonstrated with following simple example:
class A
{
public:
int operator+(int) { return 42; }
};
class B
{
public:
double operator+(double) { return 3.14; }
};
class C : public A, public B
{
};
int main()
{
C c1;
int i = 5;
double d = 10.5;
c1 + i; // Error!
c1 + d; // Error!
return 0;
}
Alex