C++ Logo


Advanced search

Re: Video Intro to WG21/D2288 C++ Designated Arguments

From: Antoine Viallon <antoine_at_[hidden]>
Date: Tue, 14 Dec 2021 15:52:58 +0100
I really do not understand your point.
What is proposed here, is a way to have designated function parameters.

I know of std::forward problems of course, but I think this is a bit unrelated, isn't it?

Have a nice day,
Antoine Viallon

Le 22 novembre 2021 21:16:13 GMT+01:00, Jason McKesson via Std-Proposals <std-proposals_at_[hidden]> a écrit :
>I think this is the first designated initializer proposal that has a
>solution to variadic template forwarding. It's a hideously ugly
>solution requiring changing template parameters on functions and
>adding a new keyword. But props for taking the problem seriously.
>I feel like the main problem with the variadic forwarding issue could
>be eliminated if `std::forward` weren't a thing. That is, if there was
>some kind of operator that *explicitly* meant "forward this
>parameter", this operator could also automatically throw in designated
>names for each argument if the forwarding operator were used in a
>place that allows designated names. Obviously if you're forwarding
>something that isn't the name of a parameter, it wouldn't work.
>So maybe this "fwdarg" keyword could be used that way. So
>`make_unique` just becomes:
>template<typename _Tp, typename... _Args>
>inline typename _MakeUniq<_Tp>::__single_object
>make_unique(_Args&&... __args)
> return unique_ptr<_Tp>(
> new _Tp(fwdarg(__args)...)
> );
>Note that a variadic parameter pack in the function declaration
>doesn't need special syntax to say "I take designated initializers
>too". That should just be how variadic parameter packs work: if you
>provide designated arguments to them, the pack captures those
>designators. The designators are used in a pack expansion *only* if
>you explicitly use the keyword `fwdarg` on the pack.
>Note also that this is just an off-the-cuff suggestion. I haven't
>thought about it enough to be sure this is a good idea, so feel free
>to change any and all of the above rules.
>But the main point is that `fwdarg` would solve two problems: the
>needlessly wordy `std::forward` and hooking designated argument names
>to function calls that want to pass them through. It even has a good
>backwards compatibility story, as you can `#define FWDARG(arg)` as
>either `std::forward<decltype(arg)>(arg)` or `fwdarg(arg)` based on
>the C++ version.
>Std-Proposals mailing list

Received on 2021-12-14 08:53:11