Pointer in c++ – Solution in Hacker Rank - hackerranksolutions8

Problem

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, of which it doesn’t have ownership.
In order to access the memory address of a variable,val, we need to prepend it with & sign. E.g., &val returns the memory address of val.
This memory address is assigned to a pointer and can be shared among various functions. E.g. 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;
}  

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Task

You have to complete the function void update(int *a,int *b), which reads two integers as argument, and sets a with the sum of them, and b with the absolute difference of them.– a` = a+b
– b` = |a-b| 

Input Format

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

Output Format

You have to print the updated value of a and b on two different lines.Note: Input/ouput will be automatically handled. You only have to complete the function described in the ‘task’ section.P.S.: Input/ouput will be automatically handled. You only have to complete the void update(int *a,int *b) function.

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Sample Input

4
5

Sample Output

9
1

Explanation

• a’ = *a – *b
• b’ = | *a – *b |

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Solution :-

#include <stdio.h>
#include<stdlib.h>

void update(int *a,int *b) 
{
    // Complete this function    
    int x, y;
    // Complete this function  
    x = *a + *b ;
    y = *a - *b ;
    *a = x;
    *b = abs(y);

}

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;
}