C++ Call by Reference: Using pointers

C++ Call by Reference: Using pointers [With Examples]

In C++ Functions article, you learned about passing arguments to a function. This method used is called passing by value because the actual value is passed.

However, there is another way of passing an argument to a function where the actual value of the argument is not passed. Instead, only the reference to that value is passed.

---

Example 1: Passing by reference without pointers

#include <iostream>
using namespace std;

// Function prototype
void swap(int&, int&);

int main()
{
    int a = 1, b = 2;
    cout << "Before swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;

    swap(a, b);

    cout << "\nAfter swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;

    return 0;
}

void swap(int& n1, int& n2) {
    int temp;
    temp = n1;
    n1 = n2;
    n2 = temp;
}

Output

Before swapping
a = 1
b = 2

After swapping
a = 2
b = 1

In main(), two integer variables a and b are defined. And those integers are passed to a function swap() by reference.

Compiler can identify this is pass by reference because function definition is void swap(int& n1, int& n2) (notice the & sign after data type).

Only the reference (address) of the variables a and b are received in the swap() function and swapping takes place in the original address of the variables.

In the swap() function, n1 and n2 are formal arguments which are actually same as variables a and b respectively.

There is another way of doing this same exact task using pointers.

---

Example 2: Passing by reference using pointers

#include <iostream>
using namespace std;

// Function prototype
void swap(int*, int*);

int main()
{
    int a = 1, b = 2;
    cout << "Before swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;

    swap(&a, &b);

    cout << "\nAfter swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;
    return 0;
}

void swap(int* n1, int* n2) {
    int temp;
    temp = *n1;
    *n1 = *n2;
    *n2 = temp;
}

The output of this example is same as before.

In this case, the address of variable is passed during function call rather than the variable itself.

swap(&a, &b); // &a is address of a and &b is address of b

Since the address is passed instead of value, dereference operator must be used to access the value stored in that address.

void swap(int* n1, int* n2) { 
 ... .. ...
}

The *n1 and *n2 gives the value stored at address n1 and n2 respectively.

Since n1 contains the address of a, anything done to *n1 changes the value of a in main() function as well. Similarly, b will have same value as *n2.