Question

If I write the following code:

#include <iostream>

using namespace std;

int main()
{
  cout << &(int &&)123 << endl;
  return 0;
}

Then g++ complains:

foo.cc: In function ‘int main()’:
foo.cc:7:20: error: taking address of xvalue (rvalue reference)

Ok, thanks to What are rvalues, lvalues, xvalues, glvalues, and prvalues? I get that an xvalue means that it’s about to “expire”, which makes sense. But now if I do this:

#include <iostream>

using namespace std;

int main()
{
  const int &x = (int &&)123;
  cout << &x << endl;
  return 0;
}

This “works” just fine and will print an address. So, I have a few questions:

1. If the value is about to expire, why can I make a reference to it? The reference won’t keep the original object alive (right?).
2. Does such a reference result in undefined behavior? E.g. because we’re referencing an object that may have been destroyed?

In general is there a way to know the lifetime of an rvalue reference?

Answer 1

Clause 12.2, paragraphs 4-5, says that the lifetime is extended in the second example

There are two contexts in which temporaries are destroyed at a different point than the end of the full-expression. …

The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:
(none of the exceptions apply here)