On Thu, Jan 5, 2023 at 7:44 PM Phil Bouchard via Std-Proposals
<std-proposals@lists.isocpp.org> 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.

--

Phil Bouchard CTO T: (819) 328-4743 E: phil@fornux.com | www.fornux.com 8 rue de la Baie | Gatineau (Qc), J8T 3H3 Canada

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