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Hackerrank Pointers in C Solution

Hackerrank Pointers in C Solution

Beeze Aal
Beeze Aal

.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

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, , we need to prepend it with  sign. E.g., &val returns the memory address of .

This memory address is assigned to a pointer and can be shared among various functions. E.g.  will assign the memory address of  to pointer . 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  and any modification to it will be reflected at the source (). void increment(int *v) {        (*v)++; }       int main() {        int a;        scanf("%d", &a);        increment(&a);        printf("%d", a);     return 0; }

Task

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

Input Format.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

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

Output Format.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

You have to print the updated value of  and , on two different lines.

Note: Input/ouput 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

Solution in c

Approach 1.

#include <stdio.h>

void update(int *a,int *b) {
    int i,j;
    i=*a;
    j=*b;
    *a=i+j;
    *b=abs(i-j);
}

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;
}

Approach 2.

#include <stdio.h>
#include <math.h>

void update(int *a,int *b) {
    int temp = *a + *b;
    *b = abs(*a - *b);
    *a = temp;
}

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;
}

Approach 3.

#include <stdio.h>

void update(int *a,int *b) { int c;
                            c=*a;
     *a=*a + *b;
     *b= abs(c - *b);
    return(*a,*b);
}

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;
}