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Re: [std-proposals] Relocation in C++

From: Edward Catmur <ecatmur_at_[hidden]>
Date: Thu, 22 Dec 2022 17:02:18 +0100
On Thu, 22 Dec 2022 at 12:17, Sebastian Wittmeier <
wittmeier_at_[hidden]> wrote:

> If the destructor is called on the reference, the original object would be
> destroyed.
>

 Then it's just as well that that isn't what happens. Instead, the lifetime
of the variable is ended in the same manner as when an automatic variable
reaches end of scope:

std::string s;
{
    auto& r = s;
} // #1
auto& q = s;
reloc q; // #2

The effect of #2 (as a discarded-value relocation expression) is precisely
the same as that of #1.

-----Urspr√ľngliche Nachricht-----
> *Von:* Edward Catmur <ecatmur_at_[hidden]>
> *Gesendet:* Do 22.12.2022 12:00
> *Betreff:* Re: [std-proposals] Relocation in C++
> *An:* std-proposals_at_[hidden];
> *CC:* Sebastian Wittmeier <wittmeier_at_[hidden]>;
> On Thu, 22 Dec 2022 at 11:53, Sebastian Wittmeier via Std-Proposals <
> std-proposals_at_[hidden]> wrote:
>
> `reloc localvar` or `reloc param` are a solution to 'undefine' variable or
> parameter names within scope or function boundaries.
>
> The same would not be possible for reference names only:
>
>
>
> MyType a{};
>
> MyType& b = a;
>
> reloc b;
>
>
>
> Would destroy and 'undefine' b, accessing a afterwards would be UB. Right?
>
>
> No, that will end the lifetime of the reference. `a` will be unaffected.
>
>
> There would be no simple/analog way to only undefine b, but keep a intact.
> That is just how references are working.
>
> OTOH, `reloc a` gives a way to undefine a variable and it would be nice to
> do the same for a reference.
>
>

Received on 2022-12-22 16:02:31