# Output of C Programs | Set 11

• Difficulty Level : Medium
• Last Updated : 27 Dec, 2016

 #includeint 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;}

In the above program, there will be recursive calls till n is not smaller than or equal to 1.

fun(5, &x)
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fun(4, fg)
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fun(3, fg)
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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

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