HOW TO TAKE MOD OF ANY NUMBER IN C++

Problem Statement

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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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

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.

P.S.: Input/ouput will be automatically handled. You only have to complete the void update(int *a,int *b) function.

Sample Input

4
5

Sample Output

9
1

Explanation

• a′=4+5=9
• b′=|4−5|=1HERE BELOW THE SOLUTION
• TWO POINTERS CAN NEVER BE ADDED BUT CAN BE SUBTRACTED
• S

  #include <stdio.h>

void update(int *a,int *b) { // Complete this function
int r1=*a+*b; int r2; if(*a<*b){ r2=*b-*a; } else r2=*a-*b; *a=r1; *b=r2; }

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;
}

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