Behavior of function overloading with const

Question

I'm working on g++ and here I have tried to overload a function by just adding const to parameter. It works fine and when it runs, it calls the function without const

1. Is this behavior specified in the C++ standard?

2. What the reason it calls the function without const

   void print(const std::string& str){std::cout << "const" << str << std::endl;}
   
   void print(std::string& str){std::cout << str << std::endl;}
   
   int main()
   {
      std::string temp = "hello";
      print(temp);
      return 0;
   }
   

Answer 1

Reference bindings are an identity category §13.3.3.1.4) but since the latter is more cv-qualified, for §13.3.3.2, the non-const is preferred (sample code from the standard):

int f(const int &);
int f(int &);

int i;
int j = f(i); // calls f(int &)

Answer 2

That is standard behavior. Any other behavior would lead to crazy behavior. In particular, the non-const function would not be callable at all.