Re: template copy constructor

From:
"Igor Tandetnik" <itandetnik@mvps.org>
Newsgroups:
microsoft.public.vc.language
Date:
Wed, 7 Oct 2009 11:53:35 -0400
Message-ID:
<uL17UZ2RKHA.1372@TK2MSFTNGP02.phx.gbl>
Vladimir Grigoriev <vlad.moscow@mail.ru> wrote:

Well, I have started from the very beginning. I have written the
auto_ptr without template functions, and it works.


That's because you don't have any constructors from "wrong" odd_ptr
type.

For constructions as the following

odd_ptr<int> pi1( new int( 10 ) );

odd_ptr<int>pi2 = pi1;

also the copy constructor is called, i.e. the compiler skips the step
of generating odd_ptr<int>( pi1 ).


odd_ptr<int>pi2 = pi1;

is equivalent to

odd_ptr<int>pi2(pi1);

because both are of the same type. From C++98 standard:

8.5p14
- If the initialization is direct-initialization, or if it is
copy-initialization where the cv-unqualified version of the source type
is the same class as, or a derived class of, the class of the
destination... [long description of direct initialization snipped]
- Otherwise (i.e., for the remaining copy-initialization cases)... [long
description of copy initialization snipped]

So, you need constructors involving unrelated types in order to
reproduce the issue.

However looking through some articles about auto_ptr I do not find
sometimes such operator as

odd_ptr & operator=( odd_ptr_ref<T> rhs ) throw()
{
reset( rhs.r.release() );
return ( *this );
}


There should be one. See DR127:

http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-defects.html#127

Is set of constructors and operators for auto_ptr predefined in C++
standard?


Yes. From C++98 20.4.5:

namespace std {
  template<class X> class auto_ptr {
    template <class Y> struct auto_ptr_ref {};
  public:
    typedef X element_type;

    // 20.4.5.1 construct/copy/destroy:
    explicit auto_ptr(X* p =0) throw();
    auto_ptr(auto_ptr&) throw();
    template<class Y> auto_ptr(auto_ptr<Y>&) throw();
    auto_ptr& operator=(auto_ptr&) throw();
    template<class Y> auto_ptr& operator=(auto_ptr<Y>&) throw();
    ~auto_ptr() throw();

    // 20.4.5.2 members:
    X& operator*() const throw();
    X* operator->() const throw();
    X* get() const throw();
    X* release() throw();
    void reset(X* p =0) throw();

    // 20.4.5.3 conversions:
    auto_ptr(auto_ptr_ref<X>) throw();
    template<class Y> operator auto_ptr_ref<Y>() throw();
    template<class Y> operator auto_ptr<Y>() throw();
  };
}

From the draft C++0x D.9:

namespace std {
  template <class Y> struct auto_ptr_ref { };

  template <class X> class auto_ptr {
  public:
    typedef X element_type;

    // D.9.1.1 construct/copy/destroy:
    explicit auto_ptr(X* p =0) throw();
    auto_ptr(auto_ptr&) throw();
    template<class Y> auto_ptr(auto_ptr<Y>&) throw();
    auto_ptr& operator=(auto_ptr&) throw();
    template<class Y> auto_ptr& operator=(auto_ptr<Y>&) throw();
    auto_ptr& operator=(auto_ptr_ref<X> r) throw();
    ~auto_ptr() throw();

    // D.9.1.2 members:
    X& operator*() const throw();
    X* operator->() const throw();
    X* get() const throw();
    X* release() throw();
    void reset(X* p =0) throw();

    // D.9.1.3 conversions:
    auto_ptr(auto_ptr_ref<X>) throw();
    template<class Y> operator auto_ptr_ref<Y>() throw();
    template<class Y> operator auto_ptr<Y>() throw();
  };

  template <> class auto_ptr<void>
  {
  public:
    typedef void element_type;
  };
}

The differences are mostly in response to DR127.

Does the above assignment operator exist in the standard
set of functions for auto_ptr


Not in C++98, but that's considered a defect. Yes in C++0x.
--
With best wishes,
    Igor Tandetnik

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