Implementing fibonacci sequence problem using iterations

Implementing fibonacci sequence problem using iterations
cpp code

#include <iostream>
#include <iomanip>
using namespace std;

//function declaration
double fibonacci(unsigned int n);

int main()
{
	int n;
	
	cout << "Enter a number: ";
	cin >> n;
	
	//An 'If' condition to prevent the user from entering -ve integers
	if (n<0){
		cout << "Please Enter a positive integer!" << endl;
		return 0;
	}
	else
		cout << "The fibonacci number is " << fibonacci(n) << endl;
			
	cout << "The fibonacci sequence is as follows:" << endl;
	
	//Print Table Heading
	cout << setw(10) << "n" << setw(10) << " | " << setw(10) << "F(n)" << endl;
    	cout << "-----------------------------------------" << endl;
	
	//A Loop to print fibonacci numbers until 'n' in a tabular form
	for(int i=0;i<=n;i++)
		cout << setw(10) << i << setw(10) << " | " << setw(10) << fibonacci(i) << endl;

	return 0;
}

//function definition
double fibonacci(unsigned int n)
{
	//n_minus_1 represents what's supposed to be F(n-1)
	//n_minus_2 represents what's supposed to be F(n-2)
	//sum represents F(n) which equals F(n-1)+F(n-2)
	double n_minus_1=0,n_minus_2=1,sum;
	
	
	switch(n)
	{
	//if n=0 then F(n)=0
	case 0:
		return 0;
	
	//if n=1 then F(n)=1
	case 1:
		return 1;
		
	//otherwise the sum (i.e F(n)) is equal to (n_minus_1)+(n_minus_2) (i.e F(n-1)+F(n-2)
	//the counter keeps on moving the variables areound 'n' times
	//so that on each time F(n-2) becomes equal to F(n-1) and F(n-1) becomes equal to F(n)
	//and F(n) is reevaluated each time in the loop
	default:
		for(int counter=0;counter<n;counter++)
		{
			sum=n_minus_1+n_minus_2;
			n_minus_2=n_minus_1;
			n_minus_1=sum ;
		}
	}
	return sum;
}