On Mon, May 30, 2022 at 12:53 PM Edward Catmur <ecatmur@googlemail.com> wrote:
On Mon, 30 May 2022 at 10:23, Arthur O'Dwyer <arthur.j.odwyer@gmail.com> wrote:
On Mon, May 30, 2022 at 11:37 AM Edward Catmur via Std-Proposals <std-proposals@lists.isocpp.org> wrote:

We expect there to exist types that are immovable and [non-trivially] relocatable.

FWIW, I (in P1144) consider that an oxymoron. A type that is immobile/immovable cannot be relocatable, because it's not movable. Immobility is for types that cannot be moved from place to place — like `std::mutex`, for example.

That's the strongest version of immovability; types that must occupy the same storage throughout their lifetime. But there are weaker versions where a class type has no copy or move constructor but is nevertheless relocatable: [e.g.]
- types that have invariants that preclude having an empty state (gsl::non_null)

Perhaps it would help to introduce terminology distinguishing the strong case from the weaker case? Say, "immobile" for the strong case and "immovable" for the weaker case? Or, just "non-semiregular" for the weaker case, if you'd prefer.

I strongly discourage making any Memmi/Derrida–style subtle distinctions between the words "immobile" and "immovable."

I suggest phrases like "types that have an `operator reloc` but no move constructor" (describing the physical situation) or "types that are conceptually movable but lack any moved-from state" (describing the conceptual situation).

If you want to permit "types that are conceptually movable but lack any moved-from state," then you need to flesh out the semantics of operations like assignment and swapping. I've seen a lot in this thread about relocating-into-a-constructor, but not much IIRC about relocating-into-an-assignment or swapping-by-relocating.

    gsl::non_null<int*> p1, p2, p3 = ...;
    gsl::non_null<int*> q1 = reloc p1;  // OK in your world
    q1 = reloc p2;  // OK??
    std::swap(q1, p3);  // OK??

Conceptually, all of these operations should be legal, right? None of them leave a gsl::non_null<int*> object in a moved-from state. Can your `operator reloc` actually handle all of them, though?  (This would go in the paper.)

The right answer to Maciej's question is: Because you expect there to exist types that are efficiently relocatable but not trivially relocatable.
One real-world example is
    struct S {
        std::any a; // non-trivially relocatable
        std::list<int> b;  // MSVC: trivially relocatable, inefficiently movable
To move-construct an `S`, you need to move the any (using its move-ctor) and then move the list (using its move-ctor, which allocates memory because MSVC uses a sentinel-node implementation).
But to relocate an `S`, you merely need to move-and-destroy the any (still using its move-ctor, presumably) and then relocate the list (using its relocation operation, which is tantamount to memcpy, and does not allocate).
Move-constructing an `S` allocates, and is non-noexcept.
Relocating an `S` doesn't need to allocate, and can be noexcept.
However, `S` is not trivially relocatable.

Maciej replies:
> Ok, but how can a type that needs custom relocation still be efficiently relocatable?
> E.g. if you need a fixup for self-reference, then relocating won't ever be as efficient as memcpy.

By "efficient relocation," I mean that relocation can be done more efficiently than move-and-destroy. My type `S` above is a concrete example.
Relocating won't ever be as efficient as memset-to-zero, either, but that doesn't matter — the first rule of programming is "Be correct." It doesn't matter how fast you do an operation if you get its semantics wrong. Trying to relocate an `S` with memcpy will produce wrong answers. Trying to relocate it with move+destroy will produce correct answers, but will be slower than the "efficient" relocation operation that I describe above.

Edward writes:
The advantage of a syntax is that you can express trivial relocation simply by defaulting it, and it Does The Right Thing if any member becomes non-trivial.

Pedantic terminology nit: Just like with any other special member, `=default` on your `operator reloc` wouldn't mean it was trivial; it would just mean it was defaulted, i.e., memberwise.
struct A { std::string s; operator reloc(A&&) = default; };  // memberwise and trivial
struct B { std::any a; operator reloc(B&&) = default; };  // memberwise and non-trivial