What is the output in a 32 bit machine with 32 bit compiler ? C#include…

2020

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

C
#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);
 }

Answer: D. 60 50 — Initially, var1 = 5 and var2 = 10. Pointers ptr1 points to var1, and ptr2 points to var2. In rer, parameters are defined as (ptr2, ptr1) but called with…

  1. A.

    60 70

  2. B.

    50 50

  3. C.

    50 60

  4. D.

    60 50

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Show answer & explanation

Correct answer: D

Initially, var1 = 5 and var2 = 10. Pointers ptr1 points to var1, and ptr2 points to var2.

In rer, parameters are defined as (ptr2, ptr1) but called with (&ptr1, &ptr2). This swaps the mapping inside the function.

The function swaps the addresses stored in main's pointers. After swapping, local **ptr1 accesses var1 and local **ptr2 accesses var2.

Calculations update var1 to 50 (5 * 10) and var2 to 60 (10 + 50). The output prints var2 then var1.

Final Output: 60 50

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