Re: How is operator+ called with rvalue reference?

From:

SG <s.gesemann@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Fri, 9 Sep 2011 04:41:51 -0700 (PDT)

Message-ID:

<86c2ab53-b6de-46ef-b79b-4805aa21af45@f41g2000yqh.googlegroups.com>

On 9 Sep., 04:45, Nephi Immortal wrote:

I read rvalue reference article. Here is the website:
http://blogs.msdn.com/b/vcblog/archive/2009/02/03/rvalue-references-c....

Unfortunately, this article is not up-to-date with respect to the
current rvalue reference rules. But the same author also made a nice
video tutorial on this topic not too long ago:

http://channel9.msdn.com/Shows/Going+Deep/C9-Lectures-Stephan-T-Lavavej-Sta=
ndard-Template-Library-STL-9-of-n

I can also recomment Scott Meyers' recent talk on this topic which is
publicly available on the net.

I wrote my code below. I can only see that rvalue reference is used
in the first parameter of the operator+ function.

int main()
{
    string s1( "01" );
    string s2( "23" );
    string s3 = s1 + s2 + "AB";
    return 0;
}

After s3 executed, first function is called:

string operator+(
        const string& _Left,
        const string& _Right)

Then, it moves in the right direction and second function is called:

string operator+(
        string&& _Left,
        const char *_Right)

Until move assignment operator is reached.

That seems about right.

Can you please be kind to demonstrate your code?
How can two functions be called like below?

string operator+(
        const char *_Left,
        string&& _Right)

and

string operator+(
        string&& _Left,
        string&& _Right)

OK, here it is:

----------8<----------

  string world();

  int main() {
    "Hello" + world(); // invokes operator+(const char*,string&&)
    world() + world(); // invokes operator+(string&&,string&&)
    return 0;
  }

----------8<----------

Cheers!
SG