Re: Overloading == as a member or as a free function

Hi,

in the following code, when calling 'test_1<T>::operator==()', the
converting ctor is invoked in order to cast a 'test_1<derived>' to
a 'test_1<base>'. However, when calling the free function
'operator==(const test_2<T>&,const test_2<T>&)' both VC, GCC4.01,
and Comeau choke, because they won't invoke the implicit conversion.
Why isn't this allowed? What am I missing?

TIA,

Schobi

--8<----8<----8<----8<----8<----8<----8<----8<----8<----8<----8<----8<--

#include <iostream>

class base {};

class derived : public base {};

template< typename T >
class test_1 {
public:
    test_1(T) {std::cout << "test_1::test_1(T)\n";}
    template< typename U >
    test_1(U) {std::cout << "test_1::test_1(U)\n";}

    void operator==(const test_1&) const {
        std::cout << "test_1::operator==(test_1<T>)\n";
    }
};

template< typename T >
class test_2 {
public:
    test_2(T) {std::cout << "test_2::test_2(T)\n";}
    template< typename U >
    test_2(U) {
        std::cout << "test_2::test_2(U)\n";
    }
};

template< typename T >
inline void operator==(const test_2<T>&,const test_2<T>&) {
    std::cout << "operator==(test_2<T>,test_2<T>)\n";
}

int main()
{
    base b;
    derived d;

    test_1<base > t1b(b);
    test_1<derived> t1d(d);
    t1b==t1d;

    test_2<base > t2b(b);
    test_2<derived> t2d(d);
    t2b==t2d;

    return 0;
}