Consider the following ANSI-C program.#include <stdio.h> int main(){ int *ptr, a, b, c; a=5; b=11; c=20; ptr=&a; *ptr=c; ptr=&c; a=*(&b); c=*ptr-a; printf("%d",c); return(0); }The output of this program is ________. (answer in integer) Note: Assume that the program compiles and runs successfully.

Let us trace the program step by step:

1. int *ptr, a, b, c; Declares an integer pointer ptr and three integer variables a, b, and c.

2. a=5; b=11; c=20;

   • $a = 5$

   • $b = 11$

   • $c = 20$

3. ptr=&a;ptr points to the address of a.

4. *ptr=c; Assigns the value of c ($20$) to the location pointed to by ptr (variable a).

   • Now, $a = 20$.

5. ptr=&c;ptr now points to the address of c.

6. a=*(&b); Dereferencing the address of b yields the value of b ($11$).

   • Now, $a = 11$.

7. c=*ptr-a; Since ptr points to c, *ptr evaluates to the current value of c ($20$).

   • $c = 20 - 11 = 9$.

8. printf("%d",c); Prints the value of c, which is 9.