> If there is no guarantee that `reloc param;` would immediately destroy a value
> parameter, would the (ugly?) implementation-defined fallback for caller-destroy
> be that - if a move assignment operator is defined - immediately the object is
> move-assigned to a temporary variable and that temporary variable is destroyed
> in turn. And after the function call, the moved-out-object is destroyed by the
> caller?

See below.

On Thu, Dec 22, 2022 at 11:38 AM Edward Catmur <ecatmur@googlemail.com> wrote:
On Thu, 22 Dec 2022 at 11:11, Sébastien Bini <sebastien.bini@gmail.com> wrote:
That makes me wonder what would happen in the following code (I don't recall we ever took a definitive decision on that point):

void do_something(std::lock_guard<std::mutex> guard)
    if (!some_test())
        reloc guard; /* attempt to release the lock early
            as the log function doesn't need it  */
        log("thread " << std::this_thread::get_id() << " failed");

If the ABI is caller-destroy then `guard` destructor will forcibly be called at function exit.
What would happen if `guard` were a non-function-parameter local variable, or the ABI callee-destroy? In other terms, should `reloc` offer any guarantee on when the destructor is called?

If we want to keep `reloc` consistent in all situations, then `reloc x;` should never call the destructor of x (which will be destroyed normally at its end of scope). If this approach is taken then everyone is dragged down because of that ABI issue which only some have and that may be resolved in the future.
This is a missed opportunity in my opinion. `reloc x;` should, when possible, call the destructor right away. That would allow developers to preemptively call the destructor of an object, without wrapping the object in an optional (or use unique_lock in our case). The language will keep track of the destruction state for us (especially is used in conditional branches), and this would no longer be a burden for the developer (which means less bugs).

I believe that `reloc src`:
  • if `src` is a local object and not a function parameter, then `src` must be left in a destructed state (either because it was passed to a relocation constructor or by a direct call to its destructor) at the end of the expression evaluation.
  • otherwise (`src` is a function parameter passed by value) then when `src` is destroyed is implementation-defined. Typically, as soon as the ABI permits.
In the last case, if a `reloc src;` statement is used and the ABI does not allow to call the destructor right-away, then compilers can still emit a warning.

I think this is too dangerous. If `src` is movable, then it's OK to move-construct a temporary and defer destruction of the moved-from object (compilers can warn on this if they feel like it). But if it is relocate-only, and the ABI is caller-destroy on that parameter, `reloc src` must be ill-formed.

Yes, I have in mind to force callee-destroy ABI for functions that take relocate-only value parameters.
But we want that to work, so if compilers want to allow relocate-only parameters to be caller-destroy, that must be a compiler extension, with the default for such parameters being callee-destroy. There should probably be an example showing this:

struct A { A() = default; A(A&&); ~A(); };
struct B { B() = default; B(B); ~B(); };
void f(A a, B b) {
    reloc a;  // may move and destroy at end
    reloc b;  // must destroy immediately

I get what you mean, but B being relocate-only, `f` will get callee-destroy ABI so in your case `reloc a` won't defer the destructor call.

Okay, things are getting confused for me so let's write things down.

`reloc src` will turn `src` into a prvalue (a temporary). If the temporary is materialized, then it will be as follows:
  1. If `src` is not a value parameter, or the ABI is callee-destroy, then the temporary is initialized using the relocation, move or copy constructor. If the last two are picked, then the destructor of the source object is called ;
  2. Otherwise, the temporary is initialized using the move or copy constructor, and the destructor of the source object is deferred until the function exits.
That being said, we don't know if the `reloc src;` statement should require the temporary materialization. Given today's rules, that would be yes. But it's really not necessary should we be in the first case: the destructor can simply be called.
If no materialisation is needed, then we can write `reloc guard;` (with guard of type `std::lock_guard`), and it will simply destroy the guard, even so lock_guard has no copy, move or reloc constructor.

I am in favor of not requiring the temporary materialization in that case. It will still be required in the second case (`src` is a value parameter and ABI is not callee-destroy).

Having said that, things flow naturally for `reloc src;` :
  • if we are in the first case and materialisation is elided, then we simply call the destructor of src ;
  • if we are in the first case and materialisation is needed, then we materialize the temporary and destruct it. It has the same effect as destroying `src` directly, except that it's suboptimized and ill-formed for immovable objects ;
  • if we are in the second case, then we materialize the temporary and destruct it. The source object destruction is deferred until the function ends as it is the only thing we can do. Compilers should emit a warning in that case IMO.
Hence, in my code snippet where I relocate a lock_guard value parameter (reloc guard;), the program is either ill-formed if the ABI is not callee-destroy or materialization is required (because lock_guard has no move or copy constructor to create a temporary), or else calls the destructor of the guard. And in:

void do_something(std::mutex& m)
    std::lock_guard guard{m};
    if (!some_test())
        reloc guard;
        log("thread " << std::this_thread_get_id() << "failed");

`reloc guard;` would either be ill-formed if materialisation is required, or will simply call the destructor of the guard.