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Re: proposal: new const-able feature ( use "const?" , or new "constable" keyword) for method/function

From: Barry Revzin <barry.revzin_at_[hidden]>
Date: Tue, 17 Dec 2019 07:07:15 -0600
 On Sun, Dec 15, 2019 at 11:51 PM jianping z <zjpwork_at_[hidden]> wrote:

> If we need both "const?" and "&&?" in a declaration/definition of a class
> method or a normal function, I think it's not difficult for compiler to
> handle them.

The question isn't so much if it's difficult for the compiler to handle.
The difficulty for the compiler probably lies in the fact that the ? can
appear arbitrarily late in a declaration. The question is more: what are
the specific rules you're proposing and how do they work?

On Mon, Dec 16, 2019 at 12:58 AM jianping z <zjpwork_at_[hidden]> wrote:

> do we really need to write "return std::move(value)" instead of simply
> "return value"? the new compiler should be able to handle it automatically
> for method with return type T&&.

Yes. The language will only implicitly move from things that are imminently
being destroyed: either local variables or function parameters that have
automatic storage duration, or (in C++20) also a few cases where the
variable being returned has rvalue reference type. None of these apply here
- if you want to move from a member variable, you have to do so explicitly.

On Mon, Dec 16, 2019 at 3:06 PM jianping z <zjpwork_at_[hidden]> wrote:

> also, for your example, do we really need following 2 member methods
> returning value of type "T&&" and "const T&&"?
> auto operator*() && -> T&& { return std::move(value); }
> auto operator*() const&& -> T const&& { return std::move(value); }
> do you have code example to show the possible usage of these 2 methods?
The first one is clear - if we have an rvalue optional we want to be able
to move from its internals instead of copying. The second one, less so.
const rvalues are a bit strange, and you can't really move from a const
rvalue anyway so there isn't performance gain there if we omit that
overload -- but the goal here is to preserve the value category of the
input. We get an rvalue in --> we get an rvalue out. The preservation
ensures, for instance, that this fails:

auto foo() -> optional<int> const;
std::ref(foo().value()); // foo() is a const rvalue, foo().value() is a int
                         // and this overload is deleted for rvalues

If we didn't provide the const rvalue overload, foo().value() would have
type int const& and we could pass it to std::ref.


Received on 2019-12-17 07:09:52