Re: How is operator+ called with rvalue reference?
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
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