Pointers in C | HackerRank Solution

Well Coderz, Today we will be solving Pointers in C HackerRank Solution.

Table of Contents

Objective

In this challenge, you will learn to implement the basic functionalities of pointers in C. A pointer in C is a way to share a memory address among different contexts (primarily functions). They are primarily used whenever a function needs to modify the content of a variable that it does not own.

In order to access the memory address of a variable, val, prepend it with & sign. For example, &val returns the memory address of val.

This memory address is assigned to a pointer and can be shared among various functions. For example, int *p = &val will assign the memory address of val to pointer p. To access the content of the memory to which the pointer points, prepend it with a *. For example, *p will return the value reflected by val and any modification to it will be reflected at the source (val).

void increment(int *v) {
        (*v)++; 
    }
      	int main() {
        int a;
        scanf("%d", &a);
        increment(&a);
        printf("%d", a);
    	return 0;      
    }

Task

Complete the function void update(int *a,int *b). It receives two integer pointers, int* a and int* b. Set the value of a to their sum, b and to their absolute difference. There is no return value, and no return statement is needed.

a’ = a+b
b’ = |a-b|

Input Format

The input will contain two integers, a and b, separated by a newline.

Output Format

Modify the two values in place and the code stub main() will print their values.

Note: Input/output will be automatically handled. You only have to complete the function described in the ‘task’ section.

Sample Input

4
5

Sample Output

9
1

Explanation

a’ = 4 + 5 = 9
b’ = | 4 – 5 | = 1

Solution – Pointers in C HackerRank Solution

#include <stdio.h>

void update(int *a,int *b) {
    // Complete this function    
     int sum,diff;
     sum = *a+*b;
     diff = abs(*a-*b);
     *a = sum;
     *b = diff;
}

int main() {
    int a, b;
    int *pa = &a, *pb = &b;
    
    scanf("%d %d", &a, &b);
    update(pa, pb);
    printf("%d\n%d", a, b);

    return 0;
}