How Pointers Are Working On Function In C Programming?

In this tutorial we are going to discuss about working of pointers on function in different aspects with examples.

In C programming, it is also possible to pass addresses in function argument. So let’s take an example:

Example: Pass Addresses To Functions

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

int main()
{
    int number1 = 5, number2 = 10;

    // address of num1 and num2 is passed
    swap( &number1, &number2);

    printf("number1 = %d\n", number1);
    printf("number2 = %d", number2);
    return 0;
}

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

When we run the program the output will be,

Output:
number1 = 10
number2 = 5

The address of number1 and number2 are passed to the swap() function using this line of code, swap(&number1, &number2).

Then in function definition part it is declaring two pointer variables like “int *n1″ and “int *n2“,

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

When the values are changes between *n1 and *n2 inside the swap() function, then the values inside the number1 and number2 are also changed in main() function.

Due to passing of addresses from main() function to swap(), then *n1 and *n2 swapped. Hence, number1 and number2 are also swapped.

Also Notice Here, swap() function is not returning anything because its return type void.
Lingaraj Senapati

Example: Passing Pointers to Functions

#include <stdio.h>

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

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

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

Here, in above program we are passing pointer variable as function argument.

So, addOne() function passing pointer q from main() as an actual argument, and in the formal argument, it’s stored in a pointer variable int* ptr.