Re: template copy constructor

From:
"Vladimir Grigoriev" <vlad.moscow@mail.ru>
Newsgroups:
microsoft.public.vc.language
Date:
Thu, 8 Oct 2009 16:58:59 +0400
Message-ID:
<u3eC9bBSKHA.1796@TK2MSFTNGP02.phx.gbl>
Now I have included some template operators
The definitions looks the following way

template <typename T>

class odd_ptr

{

private:

    template <typename U>

    struct odd_ptr_ref

    {

        odd_ptr_ref( odd_ptr<U> &rhs ): r( rhs ){}

        odd_ptr<U> &r;

    };

public:

    typedef T element_type;

    explicit odd_ptr( T *p = 0 ) throw(): ptr( p ){}

    odd_ptr( odd_ptr<T> &rhs ) throw(): ptr( rhs.release() ){}

    odd_ptr( odd_ptr_ref<T> rhs ) throw(): ptr( rhs.r.release() ){}

    ~odd_ptr() throw()

    {

        delete ptr;

    }

    odd_ptr<T> & operator=( odd_ptr<T> & rhs ) throw()

    {

        reset( rhs.release() );

        return ( *this );

    }

    template <typename U>

    odd_ptr<T> & operator=( odd_ptr<U> &rhs ) throw()

    {

        reset( rhs.release() );

        return ( *this );

    }

    odd_ptr<T> & operator=( odd_ptr_ref<T> rhs ) throw()

    {

        reset( rhs.r.release() );

        return ( *this );

    }

    template <typename U>

    operator odd_ptr_ref<U>() throw()

    {

        return ( odd_ptr_ref<U>( *this ) );

    }

    T & operator*() const throw()

    {

        return ( *ptr );

    }

    T * operator->() const throw()

    {

        return ( return ( ptr ) );

    }

    T * get() const throw()

    {

        return ( ptr );

    }

    T * release() throw()

    {

        T *p = ptr;

        ptr = 0;

        return ( p );

    }

    void reset( T *p = 0 )

    {

        if ( ptr != p ) delete ptr;

        ptr = p;

    }

private:

    T *ptr;

};

And I have defined two class for testing

class X

{

public:

    virtual ~X(){}

    virtual std::ostream & out( std::ostream &os ) const

    {

        os << "Inside X object\n";

        return ( os );

    }

};

class Y: public X

{

public:

     virtual ~Y(){}

     virtual std::ostream & out( std::ostream &os ) const

    {

        os << "Inside Y object\n";

        return ( os );

    }

};

inline std::ostream & operator<<( std::ostream &os, const X &rhs )

{

    return ( rhs.out( os ) );

}

template <typename T>

inline bool operator!( const odd_ptr<T> &rhs )

{

    return ( !rhs.get() );

}

template <typename T>

inline odd_ptr<T> f()

{

    return ( odd_ptr<T>( new T( 0 ) ) );

}

The following code is comliled well

odd_ptr<Y> py1( new Y );

std::cout << *py1 << std::endl;

odd_ptr<X> px1;

px1 = py1;

Here template <typename U> odd_ptr<T> & operator=( odd_ptr<U> & rhs )
throw() is called.

However the code

px1 = f<Y>();

is not compiled. I though that by analogy with assignment of objects of the
same types at first template <typename U> operator odd_ptr_ref<U>() will be
called and then idd_ptr<T> & operator=( odd_ptr_ref<T> &rhs ) will be called
too.

However I get an error. It seems that the compiler does not want to call the
conversion operator to convert odd_ptr<U> to odd_ptr_ref<T>. What is the
matter?

Vladimir Grigoriev

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