And how are overload resolution supposed to work with this?

```
void foo(auto&&);
void [[unimplemented]] foo(float);

int main() {
  foo(1.f); // Will it find the unimplemented foo(float) or instantiate a
speciallisation for foo(float&&)?
}

// Robin

On Wed, Mar 19, 2025, 20:01 Sebastian Wittmeier via Std-Proposals <
std-proposals_at_[hidden]> wrote:

> That is probably the best explanation.
>
>
>
> So we would potentially have 3 declarations/definitions:
>
> - An [unimplemented] declaration
>
> - An actual declaration
>
> - A definition (possibly in another TU)
>
>
> In the status quo, if the actual declaration and the definition is
> combined, the error would appear in the link stage. But the error should be
> earlier.
>
> In the status quo, if the unimplemented declaration would delete instead,
> this decision could not be overridden, e.g. in another included header.
>
> In the status quo, if the unimplemented declaration would just be left
> out, the actual declaration would not have to follow the unimplemented
> declaration, or if it does not exist, the call site would not be checked
> well,
>
>
>
> So it is kind of an unimplemented "interface", until it is declared as
> implemented.
>
>
>
> -----Ursprüngliche Nachricht-----
> *Von:* Andre Kostur via Std-Proposals <std-proposals_at_[hidden]>
> *Gesendet:* Mi 19.03.2025 16:48
> *Betreff:* Re: [std-proposals] unimplemented attribute
> *An:* std-proposals_at_[hidden];
> *CC:* Andre Kostur <andre_at_[hidden]>;
> Not that I agree with the proposal, but the scenario I'm hearing:
> 1) A cpp file includes a header with an [[unimplemented]] function
> (say the modf example above)
> 2) That cpp file then includes a header with the same function, but
> this time without the [[unimplemented]] attribute
> 3) The cpp file then calls that function.
>
> If one was to use =delete instead, #2 would be a compiler error as
> that would be redeclaring the same function. And if #2 doesn't exist,
> then the compilation could emit a diagnostic (before it got to the
> link stage). Though I'm not yet sold that simply not declaring the
> function at all isn't the better option anyway.
>
> --
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>