Determine the size of following structure

[Lil heaven]

Struct student
{ int rno ;
   char name[20] ;
   int n1, n2, total ;
}

[Banfa]

That rather depends on the platform and the packing rules in play which can be changed on some compilers.

Have you tried using the sizeof operator, it would give you an answer although it will be platform specific.

[dev7060]

Determine the size of following structure

Total size = Size of the data members + Padding size

About padding:
Padding is implementation-defined. Added between the members and/or at the end to arrange memory and avoid alignment issues. Why padding? Because the processor can read one word at a time (word size = 4 bytes in 32-bit system and 8 in 64-bit system).

#include <stdio.h>

struct student {
  int rno;
  char name[20];
  int n1, n2, total;
} obj;
int main() {
  printf("Padding = %d", sizeof(obj)-(4 * sizeof(int) + 20 * sizeof(char)));
  return 0;
}

O/P: Padding = 0

Everything is aligned (multiples of 4).

Changing to introduce padding:

#include <stdio.h>

struct student {
  int rno;
  char name[19];
  int n1, n2, total;
} obj;
int main() {
  printf("Padding = %d\n", sizeof(obj)-(4 * sizeof(int) + 19 * sizeof(char)));
  printf("%u\n", &obj.rno);
  printf("%u\n", &obj.name);
  printf("%u\n", &obj.n1);
  printf("%u\n", &obj.n2);
  printf("%u\n", &obj.total);
  return 0;
}

O/P:
Padding = 1
3281444928
3281444932
3281444952
3281444956
3281444960

The padding of 1 empty byte is added before n1. The order of the data members could change the padding and hence the structure size.