Asked by Shobhit
C
In the above program, there will be recursive calls till n is not smaller than or equal to 1.
#include<stdio.h>
int fun(int n, int *fg)
{
int t, f;
if(n <= 1)
{
*fg = 1;
return 1;
}
t = fun(n-1, fg);
f = t + *fg;
*fg = t;
return f;
}
int main( )
{
int x = 15;
printf ( "%d\n", fun (5, &x));
getchar();
return 0;
}
fun(5, &x)
\
\
fun(4, fg)
\
\
fun(3, fg)
\
\
fun(2, fg)
\
\
fun(1, fg)
fun(1, fg) does not further call fun() because n is 1 now, and it goes inside the if part. It changes value at address fg to 1, and returns 1.
Inside fun(2, fg)
t = fun(n-1, fg); --> t = 1 /* After fun(1, fg) is called, fun(2, fg) does following */ f = t + *fg; --> f = 1 + 1 (changed by fun(1, fg)) = 2 *fg = t; --> *fg = 1 return f (or return 2)Inside fun(3, fg)
t = fun(2, fg); --> t = 2 /* After fun(2, fg) is called, fun(3, fg) does following */ f = t + *fg; --> f = 2 + 1 = 3 *fg = t; --> *fg = 2 return f (or return 3)Inside fun(4, fg)
t = fun(3, fg); --> t = 3 /* After fun(3, fg) is called, fun(4, fg) does following */ f = t + *fg; --> f = 3 + 2 = 5 *fg = t; --> *fg = 3 return f (or return 5)Inside fun(5, fg)
t = fun(4, fg); --> t = 5 /* After fun(4, fg) is called, fun(5, fg) does following */ f = t + *fg; --> f = 5 + 3 = 8 *fg = t; --> *fg = 5 return f (or return 8 )Finally, value returned by fun(5, &x) is printed, so 8 is printed on the screen
