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  1. Prepare
  2. C
  3. Introduction
  4. Pointers in C
  5. Discussions

Pointers in C

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  • ahmadshahwan38
     + 0 comments

    include

    include

    void update(int *a,int *b) {

    *a= (*a + *b);
*b= abs(*a - *b- *b);

    } int main() { int a, b;

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

return 0;

    }

  • JoyceOrtiz45
     + 0 comments

    This challenge offers a great opportunity to build foundational skills and apply them to practical scenarios! 11xplay.pro

  • samarthrjain240
     + 0 comments
    #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;
}

  • raghavsmart1213
     + 0 comments

    Addition and subtraction using pointers

    I included <stdlib.h> to use abs() function. Here you can see my code snippet for your reference. If you have any idea feel free to comment.

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

void update(int *a,int *b) {
    //declaring temporary variable to store the values
    int c,d;
    c = *a +*b;  
    d = *b - *a;

    //assigning value to pointer variable
    *a = c;
    *b = d;
}

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

    //function calling
    update(pa, pb);
   
//using abs() to return positive integer according to requirement
 printf("%d\n%d", a, abs(b));

    return 0;
}

  • vinitha132415
     + 0 comments

    include

    include

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

    int main() { int a, b; int *pa = &a, *pb = &b;

    scanf("%d %d", &a, &b);
update(&a, &b);
printf("%d\n%d", a, b);

return 0;

    }