C++: rvalue reference converted to non-const lvalue-reference

2
votes

I have read this and this, but I still do not understand why the following code compiles in XCode:

void func2(string &s) {
    s = "yyyyy";
}

void func(string &&s) {
    func2(s);
}

int main() {
    func("xxxxx");
    return 0;
}

I think an rvalue reference shouldn't be converted to a non-const lvalue reference, right? In general, what's the rule of conversion between lvalue references and rvalue references? I already know that const lvalue reference can bind to rvalues, but what about rvalue references (rather than rvalues)? Thanks a lot!

1
c++c++11pass-by-referencec++14
In string &&s s is an lvalue which can bind to an lvalue reference. - nwp
I thought s is an rvalue reference. Why is it an lvalue? Is it because that it has identity and it is movable? If so, what is the difference between them? Thanks! - Yunsheng Bai
Remember if an object has a name it is a lvalue - NathanOliver

1 Answers

2
votes

The r-value reference is a reference to the original object, so converting it to a l-value reference will just make a reference to the original object.
Once a move constructor is called upon the reference, the original object should be reset to the origin state, and so does any reference to it.

This example might clarify it:

#include <iostream>
using namespace std;

int main()
{
    string s = "my string";
    string &&rval = move(s);
    cout << '"' << rval << '"' << endl; // "my string"
    cout << '"' << rval << '"' << endl; // "my string"
    cout << '"' << s << '"' << endl;    // "my string"
    string &lval = rval;
    cout << '"' << lval << '"' << endl; // "my string"
    string s2(move(rval));
    cout << '"' << rval << '"' << endl; // ""
    cout << '"' << lval << '"' << endl; // ""
    cout << '"' << s << '"' << endl;    // ""
    cout << '"' << s2 << '"' << endl;   // "my string"
    return 0;
}