Passing Pointers to Functions in C

Passing Pointers to Functions in C

Passing pointers to functions in C; Through this tutorial, you will learn how to pass pointers as arguments to functions in c programming with the help of examples.

Pointers as Function Argument in C

In C programming, Pointer as a function parameter is used to stores addresses of arguments in memory, passed during the function call. This is also known as call by reference. When a function is called by reference any change made to the reference variable will affect the original variable.

Note that – In C programming, it is also possible to pass the address of a variable to the function instead of the variable value.

It is possible to declare a pointer pointing to a function as an argument; will be declared as follows:

type (*pointer-name)(parameter);

Similarly, If a function wants to accept an address of two integer variable then the function declaration will be,

return_type function_name(int*,int*);

Here is an example :

int (*sum)();   //legal declaration of pointer to function
int *sum();     //This is not a declaration of pointer to function

A function pointer can point to a specific function when it is assigned the name of that function.

int sum(int, int);
int (*s)(int, int);
s = sum;

Here s is a pointer to a function sum. Now sum can be called using function pointer s along with providing the required argument values.

s (10, 20);

Example 1 – C Program To Swap Two Numbers using Pointers

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

int main()
{
    int num1 = 15, num2 = 20;

    // 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;
}

The output of the above c program will be:

num1 = 20
num2 = 15

Explanation of the above c program; as follows:

  • Passing the address of num1 and num2 variables into swap() function using this 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.

Example 2 – Passing Pointers to Functions

#include <stdio.h>

int sum(int x, int y)
{
    return x+y;
}

int main( )
{
    int (*fp)(int, int);
    fp = sum;
    int s = fp(10, 15);
    printf("Sum is %d", s);

    return 0;
}

The output of the above c program will be:

25

AuthorAdmin

My name is Devendra Dode. I am a full-stack developer, entrepreneur, and owner of Tutsmake.com. I like writing tutorials and tips that can help other developers. I share tutorials of PHP, Python, Javascript, JQuery, Laravel, Livewire, Codeigniter, Node JS, Express JS, Vue JS, Angular JS, React Js, MySQL, MongoDB, REST APIs, Windows, Xampp, Linux, Ubuntu, Amazon AWS, Composer, SEO, WordPress, SSL and Bootstrap from a starting stage. As well as demo example.

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