C recursion problems

Question

I have a query regarding a recursive function. This is my program.

#include <stdio.h>

int fun(int);

int main(void)
{
    int x;
    int k=3;
    x = fun(k);
    printf("%d\n",x);
    return 0;
}

int fun(int a)
{
    int f;
    if(a == 1) return (1);
    f = 2 + fun(a-1);
    return (f);
}

where I have a value K=3 in STEP-6. In STEP-7 the function fun(k) pass the value of K to the called function in STEP-11 int fun(int a). In called function fun(int a) recursion occurred 2 times that is (3,2) making the value of a=1. Which is later in STEP-14, the value of f becomes 3, because f = 2 + (fun(1)=1). In STEP-16 it get return to called function i.e fun(int a)=3. Which is supposed to print the value of x is 3, unlikely it is not. It is x =5

Answer 1

Let's check the calling sequence of fun(), shall we?

With an argument value of 3, starting from main()

• x = fun(3)
  • f = 2 + fun(2);
    • f = 2 + fun(1);

Now, let's check the return values, just in the reverse order.

1. The last call, fun(1) returns 1,
2. So the second call, fun(2), returns 2 + 1 or 3,
3. The last call, fun(3) returns 2 + 3 or 5

and that is the call made from main(). So, in main(), x gets a value of 5.

Answer 2

The evalution of fun(3) looks like this:

fun(3)
2 + fun(3-1)
2 + fun(2)
2 + 2 + fun(2-1)
2 + 2 + fun(1)
2 + 2 + 1
5

From your description, I think you have some misconceptions about scopes in C (and recursion in general). The fact that f is assigned the value 3 inside fun(2) does not mean that the value of f in the scope of fun(3) changes - they are entirely separate variables.