Using recursion to sum numbers

Question

I have just been studying the concept of recursion and I thought that I would try a simple example. In the following code, I am attempting to take the numbers: 1, 2, 3, 4, 5, an

Answer 1

Actually, I think you don't need to check case else because

public static int sum(int number){
    if(number > 0){
       return number + sum(--number);
    }
    return number; // return 0 so that's why you don't need check else condition
}

Answer 2

I always prefer to put the terminating case(s) up front so they're obvious, and I have a violent near-psychopathic hatred of "if cond then return a else return b" constructs. My choice would be (making it clear that it won't work properly for negative numbers):

static unsigned int Sum(unsigned int value) {
    if (value == 0)
        return 0;
    return value + Sum(value - 1);
}

I believe that's far more readable than a morass of braces and control flow.

Answer 3

Your terminating expression is at issue. When value == 0 (or lower), it should return a 0 rather than 1. For sake of efficiency (which, let's admit it here, obviously isn't a concern, otherwise recursion wouldn't have been used for this task), you should terminate the recursion at value == 1 and return a literal 1 to save one unnecessary level of recursion.

Answer 4

It could also be written like this:

public static int sum(int n){
    int total;

    if(n==1){
        total =1;

    }else{
        total = sum(n-1)+n;
    }
    return total;
}

Answer 5

That's because, when the value is = 0, you return 1. Then it get's added.

Sum's "else" clause should return 0.

Answer 6

You're returning 1 in the else clause. You should be returning 0:

else
{
    return 0;
}

If the value is not greater than zero, why would you return one in the first place?