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Re: [std-proposals] pass() replacement for: forward<>()

From: Phil Bouchard <boost_at_[hidden]>
Date: Thu, 5 Jan 2023 20:57:47 -0500
On 1/5/23 20:35, Jason McKesson via Std-Proposals wrote:
> On Thu, Jan 5, 2023 at 7:44 PM Phil Bouchard via Std-Proposals
> <std-proposals_at_[hidden]> wrote:
>> Greetings,
>> Please find attached a small wrapper function that could be useful as a replacement for forward<>() which uses explicit type specification (for some reason),
> There is a reason. The fact that you don't know what that reason is is
> why `pass` can't actually do what you claim. The odd thing is, your
> code doesn't even show you replacing `forward`; `::foo` still uses
> `forward`.

The forward<>() in the code is an example of the old usage.

> `std::forward` is meant to be used in perfect forwarding scenarios.
> This means you have a forwarding reference. Your example only uses
> `pass` with a non-template function. So forwarding references aren't
> around.
> Really, your `pass` is no different than using `std::move`. If I make
> a `foo_pass` template that's like `foo` except that you use `pass`
> rather than `std::forward`, then it does the wrong thing:
> https://gcc.godbolt.org/z/EMndz1xds

Oh I didn't know move() is effective for l-values and a no-op for r-values.

> So this `pass` is absolutely not a replacement for `std::forward`. It
> cannot be used for perfect forwarding.
>> shorter to type and can be used as perfect forwarding for virtual functions as well,
> Virtual functions cannot be templates and therefore cannot have
> "perfect forwarding". I think you need to take some time to learn what
> that term actually refers to.

That's fine thanks, I understand myself. So I can go ahead with a meta
universal syntax because this is really what I am looking for.

>> not only with template functions. You can use it for universal parameter types as well.
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Received on 2023-01-06 01:57:48