C++0x const RValue reference as function parameter

Question

I am trying to understand why someone would write a function that takes a const rvalue reference.

In the code example below what purpose is the cons

Answer 1

Overload resolution prefers const rvalue over const lvalue because, well, it's an rvalue and you're binding it to an rvalue reference, but you have to add const in both cases, so rvalue reference definitely preferred.

Such things are generally pointless- it's best to leave them binding to the const lvalue overloads. const rvalues don't have any real use.

Answer 2

Lvalues strongly prefer binding to lvalue references, and similarly rvalue references strongly prefer binding to rvalue references. Modifiable expressions weakly prefer binding to a non-const reference.

So when your compiler is doing overload resolution, it checks if there's an overload that takes an rvalue reference, because that's preferred strongly. In this case since the experssion is a modifiable rvalue, so the rvalue reference overload wins.

There is actually use for const rvalue references, they can be used to make sure something does not bind to an rvalue. Remember that an rvalue binds to a const lvalue reference, hence if you did:

template <typename T> void foo(const T& bar) { /* ... */ }

And called the function with:

foo(createVector());

It'd work fine. However sometimes it is desired to ensure that you can only pass lvalues to a function (this is the case for std::ref for one). You can achieve this by adding an overload:

template <typename T> void foo(const T&&) = delete;

Remember, rvalues strongly prefer binding to rvalue references, and modifiable expressions prefer weakly binding to non-const references. Since we have a const rvalue-reference, it basically means that every single rvalue will bind to this, hence if you try to pass a rvalue to foo(), your compiler will give an error. This is the only way to achieve such functionality, and thus is sometimes useful.