Is there a way to identify the const modifier of the variable in run-time?

Question

What I mean is the following question. Then I try to know the type and the constancy of the const pointer using typeinfo library, we get them both:

int* pY1 = 0;
const int* pY2 = 0;
std::cout << "pY1: " << typeid(pY1).name() << std::endl;
std::cout << "pY2: " << typeid(pY2).name() << std::endl;

Output:

pY1: int *
pY2: int const *

But then I try the following

int x1 = 0;
const int x2 = 0;   
std::cout << " x1: " << typeid(x1).name() << std::endl;
std::cout << " x2: " << typeid(x2).name() << std::endl;

an output is

x1: int
x2: int

ideone code
Is it possible to recognise the constant in the runtime? If yes, how to do this?

Answer 1

If you're using C++11 you don't need rtti at all, you can use std::is_const example :

int x1 = 0;
const int x2 = 0;
bool is_x1_const = std::is_const<decltype(x1)>::value;
bool is_x2_const = std::is_const<decltype(x2)>::value;

Old C++ version :

template<typename T> bool is_const(T) { return false;}
template<typename T> bool is_const(const T) { return true;}

Answer 2

Take the address, and you are back to your working case:

int x1 = 0;
const int x2 = 0;
std::cout << " x1: " << typeid(&x1).name( ) << std::endl;
std::cout << " x2: " << typeid(&x2).name( ) << std::endl;

Answer 3

At runtime there is no concept of constness. This is something that is used only at compile-time, which gives you the benefit of knowing constness earlier than you envision it.

If you don't have C++11 available for std::is_const, you can still copy an implementation and deduce constness with template specialization. See http://en.cppreference.com/w/cpp/types/is_const for an example implementation.

template<class T> struct is_const          : std::false_type {};
template<class T> struct is_const<const T> : std::true_type {};

You could do something similar with a function instead of a type as well, but you lose the compile-time aspect of the logic.