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What is the output of the following code?123456789101112#include <stdio.h>int main() { int x = 10; int y = 5; int *a = &x; int *b = &y; (*a)++; *b = (*b) * 2; printf("%d %d", *a, *b); return 0;}

Question

What is the output of the following code?123456789101112#include <stdio.h>int main() { int x = 10; int y = 5; int *a = &x; int *b = &y; (*a)++; *b = (*b) * 2; printf("%d %d", *a, *b); return 0;}

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Solution

The output of the code will be "11 10".

Here's the step by step explanation:

  1. The code first declares two integer variables x and y, and assigns them the values 10 and 5 respectively.

  2. Then it declares two pointers a and b, and assigns them the addresses of x and y respectively. So, a points to x and b points to y.

  3. The line (*a)++; increments the value of x by 1. So, x becomes 11.

  4. The line *b = (*b) * 2; multiplies the value of y by 2. So, y becomes 10.

  5. Finally, the printf function prints the values pointed to by a and b, which are the updated values of x and y. So, it prints "11 10".

This problem has been solved

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