References & pointers equivalence

Can anyone say that these rules below apply (I am not interested how you dereference not for that you must assign initial value when you declare reference) but if they can be used for the same think (say for pointer operations etc):



int b = 2;

int& a = b; equivalence with int *a = &b;

and

const int& a = b; equivalence with const int *const a = &b;


So if that applies can anyone provide me a solution how can I use a reference to represent?

const int *a = &b

and

int * const a = &b;


is this possible ?

Thanks

int& a = b; is equivalent to int * const a = &b; because a reference cannot change what it refers to and a const pointer cannot change what it points to.

There is no equivalent to int *a = &b; or const int* a = &b; with references.

Hi

I don’t agree with the

int &a=b; is equivalent to int * const a=&b;

as you respond me

Lets look the following program:

Code:

---

#include<iostream>
using namespace std;

int main()
{
        int w = 12;
        int b = 7;



        int* const ap = &b;

        //I can change the value
        *ap = 8;

        //Of course it fails
        //ap = &w;

        cout << *ap << '\n';

        /* From your responses
        int &a=b; is equivalent to int * const a=&b;
        */


        int& ar = b;

        //It MUST be failed but it is NOT ( I can change the reference in order to show in other variable
        // that means that it is not a const pointer because I cam change the address where it show)
        ar = w;

        cout << ar << '\n';

        //and also I can change the value
        ar++;

        cout << ar << '\n';


        //So my result is that reference is int& a = b; equivalence with int *a = &b;
        //Or maybe I do a mistake!!!!!!!!!!!!!


        return 0;
}

---

Thanks

>//It MUST be failed but it is NOT ( I can change the reference in order to show in other variable
>// that means that it is not a const pointer because I cam change the address where it show)
>ar = w;
This is where you're confused. A reference is simply another name for the object it's paired with. You're not re-assigning the reference to another object, you're changing the value of both the reference and the paired object because they're the same thing. Watch:

Code:

---

#include <iostream>

using namespace std;

int main()
{
  int a = 123;
  int b = 456;
  int& ra = a;

  cout<< a <<'\t'<< b <<'\n';
  ra = b;
  cout<< a <<'\t'<< b <<'\n';
}

---

If ra were actually paired with b, then the value of a would not change. After pairing a reference with an object, any use of the reference is exactly the same as if you replace the reference with the object it's paired with. So in the above code, every use of ra is exactly the same as if you used a instead.