What happens when you assign a literal constant to an rvalue reference?

Question

This is admittedly a nit-picky question that is mainly driven by curiosity. Suppose we have the following:

int x = 5;
int&& xref = std::move(x);
std         


        

Answer 1

A temporary is constructed, initialised from the value of the literal, and it lasts as long as the reference. You can do what you like with this object.

In terms of lifetime, this is just the same as if you'd written const int& x = 5; only, there, the fact that you're working with an automatically-created temporary object is masked because the const prevents you from proving it with a mutation.

[C++14: 8.5.3/5]: [..] If T1 is a non-class type, a temporary of type “cv1 T1” is created and copy-initialized (8.5) from the initializer expression. The reference is then bound to the temporary. [..]

Answer 2

int&& xref = 5;

... creates a temporary, initialized with 5, whose lifetime is extended to the end of the block.

The assignment

xref = 10;

changes the value of the still living temporary.