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

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:

  // 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:

  // 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:

  // 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

  // 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:

  // 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:

  // 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:

  // 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.