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What will be the output of the following code?123456789#include <stdio.h>int main() { int x = 100; int *ptr = &x; *ptr = *ptr / 3 + 5 - *ptr % 2; printf("%d", *ptr); return 0;}

Question

What will be the output of the following code?123456789#include <stdio.h>int main() { int x = 100; int *ptr = &x; *ptr = *ptr / 3 + 5 - *ptr % 2; printf("%d", *ptr); return 0;}

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Solution

The output of the code will be 33.

Here's the step-by-step explanation:

  1. The integer variable x is declared and initialized with the value 100.
  2. A pointer ptr is declared and it is assigned the address of x. So, ptr now points to x.
  3. The value at the address pointed by ptr (which is x) is updated with the result of the expression *ptr / 3 + 5 - *ptr % 2.
    • *ptr / 3 is equivalent to 100 / 3 which is 33 (since it's integer division, the fractional part is discarded).
    • *ptr % 2 is equivalent to 100 % 2 which is 0 (since 100 is an even number).
    • So, the expression becomes 33 + 5 - 0 which is 38.
  4. The printf function prints the value at the address pointed by ptr (which is now 38).

So, the output of the code will be 38.

This problem has been solved

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