How to determine programmatically if an expression is rvalue or lvalue in C++?

What's the best way to determine if an expression is a rvalue or lvalue in C++? Probably, this is not useful in practice but since I am learning rvalues and lvalues I thought it would be nice to have a function is_lvalue which returns true if the expression passed in input is a lvalue and false otherwise.

Example:

std::string a("Hello");
is_lvalue(std::string()); // false
is_lvalue(a); // true  

Most of the work is already done for you by the stdlib, you just need a function wrapper:

template <typename T>
constexpr bool is_lvalue(T&&) {
  return std::is_lvalue_reference<T>{};
}

in the case you pass a std::string lvalue then T will deduce to std::string& or const std::string&, for rvalues it will deduce to std::string

Note that Yakk's answer will return a different type, which allows for more flexibility and you should read that answer and probably use it instead.

I solved the above question using two overloaded template functions. The first takes as input a reference to a lvalue and return true. Whereas the second function uses a reference to rvalue. Then I let the compiler match the correct function depending on the expression passed as input.

Code:

#include <iostream>

template <typename T>
constexpr bool is_lvalue(T&) {
    return true;
}

template <typename T>
constexpr bool is_lvalue(T&&) {
    return false;
}

int main()
{
    std::string a = std::string("Hello");
    std::cout << "Is lValue ? " << '\n';
    std::cout << "std::string() : " << is_lvalue(std::string()) << '\n';
    std::cout << "a : " << is_lvalue(a) << '\n';
    std::cout << "a+b : " << is_lvalue(a+ std::string(" world!!! ")) << '\n';
} 

Output:

Is Lvalue ? 
std::string() : 0
a : 1
a+b : 0