Recursion problem solution not clear

[BABU Prasad G]

Hi every one, please any one can explain following program output .

#include<stdio.h>
int func(int *a, int n)
  {
    if(n ==0) return 0;
       else if(*a %2 == 0)
                  return *a+func(a+1, n-1);
           else
                  return *a-func(a+1, n-1); 

int main()
{
int a[] ={ 12, 7, 13,4,11,6};

printf(" = %d \n", func(a,6));
return;
}


output is : 15 but it is not clear for me how ?pls can any one explain in steps.

[Disch]

1) func(a,6) calls func pointing to '12', indicating there are 6 elements in the array.

2) func looks at 12, sees that it is even, and therefore does *a+func(a+1, n-1), which sums 12 with the result of the next call to func

3) next call to func is effectively func(a+1,5) which points to the '7'. func sees this is odd, so it subtracts 7-func( ...next call... )

4) next call is effectively func(a+2,4) which points to the '13'. Again this is odd, so it subtracts 15-func( ...next... )

5) next points to 4, which is even... so [code]4+func(...next...)

6) next points to 11, which is odd... so it subtracts

7) next points to 6, which is even... so it adds

8) next call has n=0, so func just returns 0

9) Step 7's call (6,even) takes the return value of 0, adds 6 to it, then returns 6

10) Step 6's call (11,odd) takes that return value of 6, subtracts it from 11, and returns 5 (11-6)

11) Step 5's call (4,even) takes that return value of 5, add's 4 to it... returning 9

12) Step 4's call (13,odd) does 13-9, returning 4

13) Step 3's call (7, odd) does 7-4, returning 3

14) Step 2's call (12,even) does 12+3, returning 15

15) Step 1 gets the final return value of 15 and prints it. Hence why your program outputs 15