[cfe-dev] deprecating copy construction and assignment

[Sean Silva]

Could someone explain why the second one is move-constructible and
move-assignable while the first one is not?

How does explicitly defaulting the move-constructor and move-assignment
cause A to *not* be move-constructible nor move-assignable, while omitting
them makes A be move-constructible and move-assignable?

--Sean Silva

On Wed, Mar 28, 2012 at 2:28 PM, Howard Hinnant <hhinnant at apple.com> wrote:

> On Mar 28, 2012, at 2:14 PM, David Blaikie wrote:
>
> > On Wed, Mar 28, 2012 at 11:08 AM, Howard Hinnant <hhinnant at apple.com>
> wrote:
> >> In a few instances the implicit definition of a defaulted copy
> constructor and copy assignment operator is deprecated in C++11.  E.g.:
> >>
> >> struct A
> >> {
> >>  // A(const A&) = default;  // deprecated
> >>
> >>  // A& operator=(const A&) = default;  // deprecated
> >>
> >>  ~A();
> >> };
> >>
> >> Should we warn when both -std=c++11 and -Wdeprecated is given?
> >
> > I believe we should, yes - just that no one's gotten around to
> > implementing this (it's been somewhere on my mental list).
> >
> > Aside: It'd be rather nice if these rules actually implemented the
> > classic "rule of three" (if you implement any of the three you should
> > implement all three - or 5 in C++11) but I don't think the standard
> > wording goes quite that far.
>
> It comes pretty close to the rule of 3:
>
> The implicitly default copy constructor is deprecated if there is a
> user-declared copy assignment or destructor.
>
> The implicitly default copy assignment is deprecated if there is a
> user-declared copy constructor or destructor.
>
> One needs to be more careful with defaulting the move members.  It is easy
> to default them and have them get implicitly deleted:
>
> struct member
> {
>    member();
>    member(const member&);
>    member& operator=(const member&);
>    ~member();
> };
>
> struct A
> {
>    member m_;
>
>    A() = default;
>    A(const A&) = default;
>     A& operator=(const A&) = default;
>     A(A&&) = default;
>    A& operator=(A&&) = default;
>    ~A() = default;
> };
>
> A is neither move constructible nor move assignable.  The following is
> probably what was intended:
>
> struct A
> {
>    member m_;
>
>    A() = default;
>    A(const A&) = default;
>     A& operator=(const A&) = default;
>     ~A() = default;
> };
>
> which, ironically, *is* move constructible and move assignable. :-)  But
> I'm not suggesting a warning for this.
>
> Howard
>
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