Re: template copy constructor

From:
"Igor Tandetnik" <itandetnik@mvps.org>
Newsgroups:
microsoft.public.vc.language
Date:
Thu, 8 Oct 2009 12:00:07 -0400
Message-ID:
<eBVipBDSKHA.1908@TK2MSFTNGP04.phx.gbl>
Vladimir Grigoriev <vlad.moscow@mail.ru> wrote:

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.


f<Y>() returns a temporary, which cannot bind to a non-const reference, so operator=( odd_ptr<U>& rhs) can't be used.

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?


Try pulling odd_ptr_ref out of odd_ptr class and put it at namespace scope. You see, odd_ptr<T>::operator= accepts
odd_ptr<T>::odd_ptr_ref<U>. But odd_ptr<U> can only convert itself to odd_ptr<U>::odd_ptr_ref<U>. These are two unrelated classes.
--
With best wishes,
    Igor Tandetnik

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