Question

Consider the below.

#include <string>
using std::string;

string middle_name () {
    return "Jaan";
}

int main ()
{
    string&& danger = middle_name();   // ?!
    return 0;
}

This doesn’t compute anything, but it compiles without error and demonstrates something that I find confusing: danger is a dangling reference, isn’t it?

Answer 1

Do rvalue references allow dangling references?

If you meant “Is it possible to create dangling rvalue references” then the answer is yes. Your example, however,

string middle_name () {
    return "Jaan";
}

int main()
{
    string&& nodanger = middle_name();   // OK.
    // The life-time of the temporary is extended
    // to the life-time of the reference.
    return 0;
}

is perfectly fine. The same rule applies here that makes this example (article by Herb Sutter) safe as well. If you initialize a reference with a pure rvalue, the life-time of the tempoary object gets extended to the life-time of the reference. You can still produce dangling references, though. For example, this is not safe anymore:

int main()
{
    string&& danger = std::move(middle_name());  // dangling reference !
    return 0;
}

Because std::move returns a string&& (which is not a pure rvalue) the rule that extends the temporary’s life-time doesn’t apply. Here, std::move returns a so-called xvalue. An xvalue is just an unnamed rvalue reference. As such it could refer to anything and it is basically impossible to guess what a returned reference refers to without looking at the function’s implementation.