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

[Hans Åberg]

> On 15 Sep 2018, at 21:25, Frank Heckenbach <address@hidden> wrote:
> 
> 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. Move operators were originally designed to 
avoid copying in returns.

>> Even in view of copy elision, default in C++17 [1], this would be safe, 
>> because one cannot move an already moved object by mistake.
> 
> Why not?

Because it breaks the execution path, so one cannot apply it twice to the same 
value.

>> As the point is breaking out of the execution path, one might use your 
>> suggestion of a special operator in combination with an immediately 
>> following break in the action switch statement. So writing say
>>  $$$(make_unique<foo>($1, $2, $3));
>> translates into
>>  $$ = std::move(make_unique<foo>($1, $2, $3));
>>  break; 
> 
> What if I want to write:
> 
>  $$ = make_unique <foo> ($1, $2, $3);
>  print ($$);

Then you can't use the proposed $$$, but has to use the old syntax.