Different ways to use Const with Reference to a Pointer in C++

Last Updated : 12 Jul, 2025
Before moving forward with using const with Reference to a Pointers, let us first see what they are one by one:
  • Pointers are used to store the address of variables or a memory location. A variable can be declared as a pointer by putting ‘*’ in the declaration. 
    datatype *var_name;
    Example: CPP 
    // C++ program to
// demonstrate a Pointer

#include <iostream>
using namespace std;

int main()
{

    // Variable
    int i = 10;

    // Pointer to i
    int* ptr_i = &i;

    cout << *ptr_i;

    return 0;
}
    Output: 
    10
  • Reference: When a variable is declared as a reference, it becomes an alternative name for an existing variable. A variable can be declared as a reference by putting ‘&’ in the declaration. 
    datatype &var_name;
    Example: CPP 
    // C++ program to
// demonstrate a Reference

#include <iostream>
using namespace std;

int main()
{

    // Variable
    int i = 10;

    // Reference to i.
    int& ref = i;

    // Value of i is now
    // changed to 20
    ref = 20;

    cout << i;

    return 0;
}
    Output: 
    20
  • References to pointers is a modifiable value that's used same as a normal pointer. 
    datatype *&var_name;
    Example 1: CPP 
    // C++ program to demonstrate
// References to pointers

#include <iostream>
using namespace std;

int main()
{

    // Variable
    int i = 10;

    // Pointer to i
    int* ptr_i = &i;

    // Reference to a Pointer ptr_i
    int*& ptr_ref = ptr_i;

    cout << *ptr_ref;

    return 0;
}
    Output: 
    10
    Here ptr_ref is a reference to the pointer ptr_i which points to variable 'i'. Thus printing value at ptr_ref gives the value of 'i', which is 10. Example 2: Now let us try to change the address represented by a Reference to a Pointer CPP 
    // C++ program to demonstrate
// References to pointers

#include <iostream>
using namespace std;

int main()
{

    // Variable
    int i = 10;
    int j = 5;

    // Pointer to i
    int* ptr = &i;

    // Reference to a Pointer ptr
    int*& ptr_ref = ptr;

    // Trying to change the reference
    // to Pointer ptr_ref to address of j
    ptr_ref = &j;

    cout << *ptr;

    return 0;
}
    Output: 
    5
    Here it prints 5, because the value of j is 5 and we changed ptr_ref to point to j. Now as ptr_ref is a reference to pointer ptr, ptr now points to j. Thus we get the output we expected to see.
  • Const Reference to a pointer is a non-modifiable value that's used same as a const pointer. 
    datatype* const &var_name;
    Example 1: CPP 
    // C++ program to demonstrate
// References to pointers

#include <iostream>
using namespace std;

int main()
{

    // Variable
    int i = 10;
    int j = 5;

    // Pointer to i
    int* ptr = &i;

    // Const Reference to a Pointer
    int* const& ptr_ref = ptr;

    // Trying to change the const reference
    // to Pointer ptr_ref to address of j
    ptr_ref = &j;

    cout << *ptr;

    return 0;
}
    Compilation Error: 
    In function 'int main()':
prog.cpp:23:13: error: assignment of read-only reference 'ptr_ref'
     ptr_ref = &j;
             ^
    Here we get a compile-time error as it is a const reference to a pointer thus we are not allowed to reassign it. Example 2: CPP 
    // C++ program to demonstrate
// References to pointers

#include <iostream>
using namespace std;

int main()
{

    // Variable
    int i = 10;
    int j = 5;

    // Pointer to i
    int* ptr = &i;

    // Const Reference to a Pointer
    int* const& ptr_ref = ptr;

    // Trying to change the reference
    // to Pointer ptr_ref
    *ptr_ref = 100;

    cout << *ptr;

    return 0;
}
    Output: 
    100
    It prints 100 as it is not a reference to a pointer of a const. Why Example 2 didn't throw a Compile-time error when Example 1 did?
    • In Example 1, the ptr_ref is a const reference to a pointer to int, and we are trying to change the value of ptr_ref. So the compiler throws a Compile time error, as we are trying to modify a constant value.
    • In Example 2, the ptr_ref is a const reference to a pointer to int, and we are trying to change the value of *ptr_ref, which means we are changing the value of int to which the pointer is pointing, and not the const reference of a pointer. So the compiler doesn't throw any error and the pointer now points to a value 100. Therefore the int is not a constant here, but the pointer is. As a result, the value of int changed to 100.
  • Reference to a Const Pointer is a reference to a constant pointer. 
    datatype const *&var_name;
    Example: CPP 
    // C++ program to demonstrate
// References to pointers

#include <iostream>
using namespace std;

int main()
{

    // Variable
    int i = 10;
    int j = 5;

    // Const Pointer to i
    int const* ptr = &i;

    // Reference to a Const Pointer
    int const*& ptr_ref = ptr;

    // Trying to change the value of the pointer
    // ptr with help of its reference ptr_ref
    *ptr_ref = 124;

    cout << *ptr;
    return 0;
}
Compilation Error: 
In function 'int main()':
prog.cpp:23:14: error: assignment of read-only location '* ptr_ref'
     *ptr_ref = 124;
              ^
Here again we get compile time error. This is because here the compiler says to declare ptr_ref as reference to pointer to const int. So we are not allowed to change the value of i.
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