Can't print Fibonacci series

Question

I was writing a small snippet to get a Fibonacci number sequence depending on the user input. If the user supplies 4 as an input, it should return him the first N members of the Fibonacci sequence.

#include <iostream>
using namespace std;

int main (){

    int a = 0;
    int b = 1;
    int c;
    int n = 3;
    n -= 2;

    if (n == 1){
        cout << a << endl;
    } else {
        cout << a << b << endl;

        for (int i=0;i<n;i++){
            c = b + a;
            cout << c << endl;
            a = b;
            b = c;
        }
    }

}

However, I end up getting a 0 as an output for whatever number I supply. I have this working in PHP and I kinda miss where I've blundered. I guess I don't actually render input and output properly.

Answer 1

Try this which would give you the list you needed.

#include <iostream>
using namespace std;


int fib(int num){
    int ans;
    if (num >2) {
       ans = fib(num-1) + fib(num-2);
    }
    else
       ans = 1;
    return ans;
}

int main()
{
    int num, x=1;
    cin >> num;
    while (num >= x) {
       cout << fib(x) <<" ";
       x++;
    }
    return 0;
}

Answer 2

The sequence you want is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, ...

As I pointed out in a comment to another answer, your code does not produce a correct Fibonacci sequence. F(3) isn't the problem with your code; the problem is that you get confused between all the variables, a, b, c and use them to mean different things at once.

You also incorrectly decrement n: your code does it in the wrong place, and even if you move it to the right place, it wouldn't help as the operation would make n go negative.

---

Your existing Code

Let's walk through your code a bit:

int a = 0;
int b = 1;
int c;
int n = 3;
n -= 2;

Well, this is weird. We set n to 3 then immediately subtract 2, making it 1. This means that if you try to set n to 0, 1, or 2 you end up with n being a negative number. If you set it to 3, you end up with n being 1.

if (n == 1){
    cout << a << endl;
} 

We're in trouble right here. Remember that you subtract 2 from n which means that for n==3 you will return whatever is in a which is wrong. But even if you meant this to special-case F(1) that code is still wrong because F(1)=1.

else {
    cout << a << b << endl;

    for (int i=0;i<n;i++){

Remember, that we can get here with n zero or negative. Obviously in the case of n <= 0 this loop will never execute, so c will never be printed.

        c = b + a;
        cout << c << endl;

Here, we seem to calculate and output the next Fibonacci number by adding the two previous numbers. This should be fine.

        a = b;
        b = c;

And here, we keep the new Fibonacci number and its predecessor for the next loop iteration, if any.

The problems with this code are, of course, fixable. But the problem is that the existing code is confusing. It outputs all sorts of different values, and it's unclear what variable is supposed to represent.

Looking at this problem, your first instinct would be to make a function which accepts as input a number n and returns F(n) - you could call it fib or somesuch.

---

Reworking this

So, how to go about writing such a function? Here's a simple recursive implementation that you can use:

int fib(int n)
{
    if ((n == 0) || (n == 1))
        return n;

    return fib(n-1) + fib(n-2);
}

Notice how this function is short, sweet and to the point. There's no need for a ton of variables, no need for complicated control structures or storing state. It almost reads like a text-based description of the Fibonacci algorithm.

Of course, it's not super-efficient and ends up redoing a lot of work. That's a legitimate criticism, but it's unlikely that there performance considerations here.

Still, perhaps you just don't like recursion. Many people think of recursion as a dirty word, and avoid it with a passion. So how about a non-recursive implementation instead? It's possible, but it's a bit more difficult to understand.

int fib (int n)
{
    /* F(0) = 0 */
    if (n == 0)
        return 0;

    int a = 0;
    int b = 1;

    for (int i = 2; i < n; i++)
    {
        int c = a + b;
        a = b;
        b = c;
    }

    /* F(n) = F(n-2) + F(n-1) */
    return a + b;
}

This is a little bit more efficient and not that much more difficult to understand.

I hope that this helped.

Answer 3

int a =0;    
int n = 3;
n -= 2;

if (n == 1){
    cout << a << endl;
}

You have n equal to 3, you subtract 2, thus n equal to 1, so, you enter the if body and output a, which is zero.

[EDIT]

You don't seem to get any input -as stated in a comment- in your program (you could use std::cin or std::getline() for this), but you probably mean that you have the input hard-coded, by changing the value of n by hand.

You may want to check how the Fibonacci series program is expected to work:

1. Fib. at Rosseta page.
2. Fib. with recursion
3. Non-recursive Fib.

After reading the links I provided above, you should be able to see that your code should be changed to this:

#include <iostream>
using namespace std;

int main (){

    int a = 1;
    int b = 0;
    int c;
    int n = 10; // "input" is 10

    if (n == 0 || n == 1) { // 0 and 1 case
        cout << n << endl;
    } else {
        for (int i = 2; i <= n; ++i) { // here you want to reach n
            c = a + b;
            b = a;
            a = c;
        }
        cout << c << endl;
    }

    return 0;
}

However, the code above outputs only the result. You should slightly modify it to get the terms of the sequence, but I'll leave you have some fun too.

In order to really let the user input the number, change:

int n = 10;

to

int n;
std::cout << "Please, input.\n";
std::cin >> n;

However, letting user inputting must be followed by validation of the input. You see users can, by accident or not, provide input in your program, that can cause undefined behaviour.