What is the use of const overloading in C++?

Question

In C++, a function\'s signature depends partly on whether or not it\'s const. This means that a class can have two member functions with identical signatures except that on

Answer 1

#include 
using namespace std;
class base
{

public:
void fun() const
{
    cout<<"have fun";
}
void fun()
{
    cout<<"non const";
}

};
int main()
{
    base b1;
    b1.fun(); //does not give error
    return 0;
}

Here compiler won't give any error, because in case of const functions compiler converts this pointer to const this*. this third argument separates these two functions.