Comments on: Move It With Rvalue References

By: Steven Watanabe

Steven Watanabe — Wed, 28 Oct 2009 03:36:41 +0000

Fixed. Thanks for pointing it out.

By: Christoph Heindl

Christoph Heindl — Sat, 17 Oct 2009 07:07:24 +0000

The link to the third installment is dead. It should refer to

http://cpp-next.com/archive/2009/09/making-your-next-move/

I guess.

By: Niels Dekker

Niels Dekker — Wed, 30 Sep 2009 21:46:37 +0000

I'm glad to see that your issue on this subject is included with the pre-Santa-Cruz mailing: Core Issue #964, Incorrect description of when the lvalue-to-rvalue conversion applies

By: Niels Dekker

Niels Dekker — Tue, 22 Sep 2009 12:52:33 +0000

Doug Gregor wrote: The reference to “x” is an lvalue, but it is converted to an rvalue to perform the multiplication.

Thanks, Doug. But do you mean that if [basic.lval]/7 would not have been there, one would have to convert an lvalue argument of an int multiplication explicitly? As follows?

int doubleMe(int x) { return 2 * std::move( x ) ; }

By: Doug Gregor

Doug Gregor — Tue, 22 Sep 2009 04:54:42 +0000

Niels Dekker: You’re welcome, Doug. I read your proposed change of [basic.lval], which would make sure that that particular sentence does not apply when binding an rvalue reference. Cool! Just for my understanding: In what other context would an lvalue still be converted implicitly to an rvalue?

It happens a lot with built-in types. For example, given:

int doubleMe(int x) { return 2 * x; }

The reference to "x" is an lvalue, but it is converted to an rvalue to perform the multiplication.

By: Dave Abrahams

Dave Abrahams — Thu, 17 Sep 2009 05:01:17 +0000

Dave Abrahams: Hi Nasos, The next article in this series will have a chart of binding/overloading rules that can help answer your question

Actually I realized that section belonged in this article, so I added it as an update. Search for "Binding and Overloading". Cheers!

By: Niels Dekker

Niels Dekker — Tue, 15 Sep 2009 14:44:48 +0000

You’re welcome, Doug. I read your proposed change of [basic.lval], which would make sure that that particular sentence does not apply when binding an rvalue reference. Cool! Just for my understanding: In what other context would an lvalue still be converted implicitly to an rvalue?

By: Pavel Shevaev

Pavel Shevaev — Tue, 15 Sep 2009 08:58:42 +0000

Oh, can’t wait too see it, thanks in advance!

By: Dave Abrahams

Dave Abrahams — Tue, 15 Sep 2009 07:55:09 +0000

Hi Pavel! A brief survey of C++03 move emulation is in already in the queue.

By: Dave Abrahams

Dave Abrahams — Tue, 15 Sep 2009 07:53:26 +0000

Yes, c---being an lvalue---will be copied to create f's parameter a, and the compiler is allowed to elide all other copies in this code.