gcc-help@gcc.gnu.org

[Panos C. C.]

I get a very weird error in a simple and pretty straightforward piece of code involving 2 templates. I've found a workaround but its ugly and probably unsafe. Does anyone know what is wrong with the following code. Intel C++ compiler also fails.


Error:

../main.cpp: In function ‘int main()’:
../main.cpp:54: error: no match for ‘operator=’ in ‘niaou = bs.RsrcContainer<Type>::foo [with Type = B]()’
../main.cpp:19: note: candidates are: RsrcPtr<Type>& RsrcPtr<Type>::operator=(RsrcPtr<Type>&) [with Type = B]


Code:

--------------------------------------------------------------------------------
template<typename Type>
class RsrcPtr
{
    public:
        explicit RsrcPtr(Type* p_ = NULL):
            p(p_)
        {}

        RsrcPtr<Type>& operator=(RsrcPtr<Type> &a)
        {
            p = a.p;
            a.p = NULL;
        }

    private:
        Type* p;
};


template<typename Type>
class RsrcContainer
{
    public:
        RsrcPtr<Type> foo()
        {
            RsrcPtr<Type> tmp(new Type);
            return tmp;
        }
};


class B
{
    public:
        int x;
};


int main()
{
    RsrcContainer<B> bs;
    RsrcPtr<B> niaou;
    niaou = bs.foo();
    return 0;
}
--------------------------------------------------------------------------------

If I replace operator= with this:

template<Type1>
RsrcPtr<Type1>& operator=(RsrcPtr<Type1> &a)
{
    p = a.p;
    a.p = NULL;
}


I get no error (actually auto_ptr does the same) but its not logical.
 		 	   		  
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