Question

Consider the program below:
# include <stdio.h>
int fun (int n, int *f_p) {
int t, f;
if(n< =1){
*f_p = 1;
return 1;
}
t = fun(n-1, f_p);
f = t +* f_p;
*f_p = t:
return f ;
}
int main()
{
int x = 15;
printf (“% d\n”, fun (5, &x)) ;
return 0;
The value printed is

• A. 6
• B. 14
• C. 8
• D. 15

Answer 1

The correct option is C 8

x = 15
fun (5, &x) let &x = 100
$\downarrow$
fun (5, 100)
$\downarrow$ t =5, f = 5 + 3, *f_p =5
return 8
fun (4, 100)
$\downarrow$ t = 3, f = 3 + 2, *f_p = 3
return 5
fun (3, 100)
$\downarrow$ t = 2, f = 2 + 1, *f_p = 2
return 3
fun (2, 100)
$\downarrow$ t = 1, f = 1 + -1, *f_p = 1
return 2
fun (1, 100)
*f_p = 1
return 1
So fun (5, & x) will return 8 and it will be printed.