C Call by Reference: Using pointers

In this tutorial, you’ll learn to pass addresses as arguments to the function with the help of examples. This procedure is known as call by reference.

In C programming, it is also possible to pass addresses as arguments to functions.

To acknowledge these addresses in the capacity definition, we can utilize pointers. This is on the grounds that pointers are used to store addresses. Let’s take an example:

Example: Call by reference

#include <stdio.h>
void swap(int *n1, int *n2);

int main()
{
    int num1 = 5, num2 = 10;

    // address of num1 and num2 is passed
    swap( &num1, &num2);

    printf("num1 = %d\n", num1);
    printf("num2 = %d", num2);
    return 0;
}

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

When you run the program, the output will be:

num1 = 10
num2 = 5

The address of num1 and num2 are passed to the swap() function using swap(&num1, &num2);.

Pointers n1 and n2 accept these arguments in the function definition.

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

When *n1 and *n2 are changed inside the swap() function, num1 and num2 inside the main() function are also changed.

Inside the swap() function, *n1 and *n2 swapped. Hence, num1 and num2 are also swapped.

Notice that, swap() is not returning anything; its return type is void.

This technique is known as call by reference in C programming.

Example 2: Passing Pointers to Functions

#include <stdio.h>

void addOne(int* ptr) {
  (*ptr)++; // adding 1 to *ptr
}

int main()
{
  int* p, i = 10;
  p = &i;
  addOne(p);

  printf("%d", *p); // 11
  return 0;
}

Here, the value stored at p, *p, is 10 initially.

We then passed the pointer p to the addOne() function. The ptr pointer gets this address in the addOne() function.

Inside the function, we increased the value stored at ptr by 1 using (*ptr)++;. Since ptr and p pointers both have the same address, *p inside main() is also 11.

Please feel free to give your comment if you face any difficulty here.