ISRO | ISRO CS 2020 | Question 26

What is the output in a 32 bit machine with 32 bit compiler ?

#include <stdio.h>

rer(int **ptr2, int **ptr1)
 {
  int* ii;
  ii = *ptr2;
  *ptr2 = *ptr1;
  *ptr1 = ii;
  **ptr1 *= **ptr2;
  **ptr2 += **ptr1;
 }
void main( )
 {
   int var1 = 5, var2 = 10;
   int *ptr1 = &var1, *ptr2 = &var2;
   rer(&ptr1, &ptr2);
   printf(“%d %d “, var2, var1);
 }

(A) 60 70
(B) 50 50
(C) 50 60
(D) 60 50


Answer: (D)

Explanation: Under rer functions:

ptr2 = 3000 ptr1 = 4000

ii = 1000

*(3000) = *(4000)
i.e., outer ptr1 = 2000

*(4000) = 1000
i.e., outer ptr2 = 1000

**ptr1 *= **ptr2
will convert var1 = 50

**ptr2 += **ptr1
will convert var2 = 60 

Hence, 60 50 gets printed.

#include <stdio.h>

int rer(int **ptr2, int **ptr1)
 {
  int* ii;
  ii = *ptr2;
  *ptr2 = *ptr1;
  *ptr1 = ii;
  **ptr1 *= **ptr2;
  **ptr2 += **ptr1;
 } ;

int main( )
 {
   int var1 = 5, var2 = 10;
   int *ptr1 = &var1, *ptr2 = &var2;
   rer(&ptr1, &ptr2);
   printf("%d %d ", var2, var1);
 }

Option (D) is correct.

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