Be mindful though that a prvalue is different from an xvalue. Copy elision
and direct initialization cannot happen with calls to std::move, whereas it
is permitted with prvalue. If you have a source to source compiler you will
have to take that into account to not break code, and you will have to only
generate std::move when the expression would be an lvalue.

Here's an example of a non-copiable non-movable type passed in parameter by
value:

    void do_thing_with_mutex(std::mutex);

    do_thing_with_mutex(std::mutex{});

If you call std::move on that prvalue, the code will be ill-formed since
the compiler would be forced to call the move constructor.

It may look like details, but those are important details because you might
accidentally propose things that will break people and hinder performance.

Em sáb., 7 de jan. de 2023 16:44, Phil Bouchard via Std-Proposals <
std-proposals_at_[hidden]> escreveu:

>
>
> On 1/7/23 03:35, Peter Olsson via Std-Proposals wrote:
> >> a.foo(move(list<int>{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}));
> >
> >> foo(move(ostream()));
> >
> > Do you realize that using std::move on a temporary like this is pretty
> > pointless because it's already an rvalue?
>
> I was referring to the automatic generated code by the source-to-source
> compiler. Whether it's a r-value and l-value, a move() will be generated.
>
>
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