Re: Automatially move from $n (was: C++11 move semantics)

[Frank Heckenbach]

Hans Åberg wrote:

> >>>> The idea would be to write something equivalent to
> >>>> return make_unique<foo>($1, $2, $3);
> >>>> and the Bison writes something like
> >>>> $$ = std::move(action_k(...return make_unique<foo>($1, $2, $3);...))
> >>> 
> >>> I don't follow you. What is action_k, and how would that cause
> >>> moving from $1 etc.?
> >> 
> >> Action k in the switch statement.
> > 
> > Huh? I really don't get what your proposed syntax is supposed to
> > mean. Is action_k supposed to be a lambda (what else could appear in
> > an expression and contain a statement inside)? What would it do?
> 
> Just produce an r-value.

Again:

- make_unique already produces an rvalue

- (I'll ignore the "...return", since you didn't comment on it, I
  assume it's a typo)

- Then, you say, action_k produces an rvalue, from an rvalue?

- Finally, std::move takes this rvalue and turns it into an rvalue
  (because that's what std::move does).

Do you want a triple-r-value?

Sorry if I'm a bit cynical meanwhile, but I said I don't follow what
you intend to do, so it would be nice to explain it with something
more than half a sentence, really.

> >> Move operators were originally designed to avoid copying in returns.
> > 
> > I don't know if this was so or not originally, but I'm talking about
> > moving arguments, not return values. That's what I've been saying
> > the whole time, including the thread subject! Moving the return
> > value is no big problem most of the time: "$$ = make_unique ..."
> > works without any std::move because a function result(*) is
> > automatically an rvalue.
> 
> The idea is to create an r-value situation, which then translates into a move 
> assignment.

Again, make_unique already "creates an rvalue situation".

And, also again, how does any of that help with moving from $1, $2
etc. which is the actual topic?

Regards,
Frank