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What will be the output for the following code?12345678910111213#include <stdio.h>void setZero(int *num) { *num = 0;}int main() { int i = 10; int *ptr = &i; setZero(&(*ptr)); printf("%d", *ptr); return 0;}

Question

What will be the output for the following code?12345678910111213#include <stdio.h>void setZero(int *num) { *num = 0;}int main() { int i = 10; int *ptr = &i; setZero(&(*ptr)); printf("%d", *ptr); return 0;}

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Solution

The output of the code will be 0.

Here's the step by step explanation:

  1. The main function is called. Inside main, an integer i is declared and initialized with the value 10.

  2. A pointer to an integer ptr is declared and it is initialized with the address of i. So, ptr is pointing to i.

  3. The setZero function is called with the address of i (which is what ptr is holding).

  4. Inside the setZero function, the value at the address that the pointer num is holding (which is the address of i) is set to 0. This changes the value of i to 0.

  5. Back in the main function, the value at the address that ptr is holding (which is the value of i) is printed. Since i was set to 0 in the setZero function, 0 is printed.

  6. The main function returns 0, signaling successful termination of the program.

This problem has been solved

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