Is it possible to call a function which takes a const rvalue as a parameter?

Question

Recently i accidently programmed the following:

void someFunction(const SomeClass&& value){
    //Some Code
}

I didn't mean to add the 'const' keyword to the value parameter. However, this made me think: is it even possible to somehow call this function, if an equivalent function without const exists? If yes, does this ever make sense and is there some practical use to it?

Answer 1

This code:

#include <iostream>
using namespace std;

struct s{};     

void someFunction(s& value){ std::cout << "ref" << std::endl; }

void someFunction(const s&& value){ std::cout << "const" << std::endl; }

void someFunction(s&& value){ std::cout << "non const" << std::endl; }

const s foo() { return s(); }
s bar() { return s(); }


int main() {
    someFunction(bar());
    someFunction(foo());
    return 0;
}

Prints

non const
const

For some explanation (that I cannot provide ;) I refer you to this site. The most relevant quote is:

[...] This makes const rvalue references pretty useless. Think about it: if all we need is a non-modifiable reference to an object (rvalue or lvalue), then a const lvalue reference does the job perfectly. The only situation where we want to single out rvalues is if we want to move them. And in that case we need a modifiable reference.

The only use case I could imagine is to use someFunction to detect whether a second function returns a const or a non-const, but I guess there are easier ways to find that out.

As a sidenote: When you replace s with int in above code it prints:

non const 
non const

and that is because (quoting from the same site, comment from pizer):

There are no const prvalues of scalar types. For a const rvalue you either need an Xvalue and/or a class-type object. [...]