#include <stdio.h>int main(void) { int i = 11; int *ptr1 = &i; int **ptr2 = &ptr1; int ***ptr3 = &ptr2; printf("%p, %p, %p, %p\n", ptr1, ptr2, ptr3, &ptr3); return 0;}If the program output is (where _ is part of the address that is not shown):0x_0b4, 0x_0b8, 0x_0c0, 0x_0c8At address 0x_0b4 is stored 11 .At address 0x_0c8 is stored 200 .

The program you've written in C language is creating pointers to an integer and then pointers to those pointers. Here's a step-by-step explanation:

1. int i = 11; This line declares an integer i and assigns it the value 11.

2. int *ptr1 = &i; This line declares a pointer ptr1 to an integer and assigns it the address of i. So, ptr1 points to i.

3. int **ptr2 = &ptr1; This line declares a pointer ptr2 to a pointer to an integer and assigns it the address of ptr1. So, ptr2 points to ptr1.

4. int ***ptr3 = &ptr2; This line declares a pointer ptr3 to a pointer to a pointer to an integer and assigns it the address of ptr2. So, ptr3 points to ptr2.

5. printf("%p, %p, %p, %p\n", ptr1, ptr2, ptr3, &ptr3); This line prints the addresses stored in ptr1, ptr2, ptr3, and the address of ptr3 itself.

The output of the program shows the addresses of i, ptr1, ptr2, and ptr3 respectively. The addresses will vary each time the program is run because the operating system assigns these addresses dynamically.

The statement "At address 0x_0b4 is stored 11" means that the integer i with value 11 is stored at the address 0x_0b4.

The statement "At address 0x_0c8 is stored 200" seems to be incorrect or unrelated to the given program because there is no such value 200 in the program. The address 0x_0c8 is the address of ptr3 itself, not the value it points to.