what does int numbers[n+2]; statement do? - c++

Question

#include<iostream>
int fastFibonacci(int n)
{
int numbers[n+2]; // int numbers[n].
numbers[0] = 0;
numbers[1] = 1;
for (int i = 2; i <= n; i++)
{
numbers[i] = numbers[i - 1] + numbers[i - 2];
}
return numbers[n];
}
int main() {
int n;
std::cout << "Enter a Number";
std::cin >> n;
int result = fastFibonacci(n);
std::cout << result << "\n";
return 0;
}
in this code when i enter input 0 or 1 get correct answer. But the problem is that when i replace int numbers[n+2]; with the commented part it start giving me wrong answer when input is 0 or 1. why? anyone please explain me.

Answer 1

In this function
int fastFibonacci(int n)
{
int numbers[n+2]; // int numbers[n].
numbers[0] = 0;
numbers[1] = 1;
for (int i = 2; i <= n; i++)
{
numbers[i] = numbers[i - 1] + numbers[i - 2];
}
return numbers[n];
}
there is used a variable length array with n + 2 elements declared in this line
int numbers[n+2]; // int numbers[n].
Variable length arrays is not a standard C++ feature. It can be implemented as own language extension of a C++ compiler.
Using the variable length array makes the function very unsafe because there can occur a stack overflow.
As within the function there is explicitly used two elements of the array
numbers[0] = 0;
numbers[1] = 1;
then the array shall have at least two elements even when the parameter has a value less than 2.
To calculate the n-th Fibonacci number there is no need to declare an array of such a size.
Apart from this the function argument shall have an unsigned integer type. Otherwise the function can invoke undefined behavior if the user passes a negative number.
Also for big values of n there can be an integer overflow for the type int.
The function can be implemented in various ways.
Here is one of possible its implementations.
#include <iostream>
#include <functional>
unsigned long long fibonacci( unsigned int n )
{
unsigned long long a[] = { 0, 1 };
while ( n-- )
{
a[1] += std::exchange( a[0], a[1] );
}
return a[0];
}
int main()
{
const unsigned int N = 10;
for ( unsigned int i = 0; i < N; i++ )
{
std::cout << i << ": " << fibonacci( i ) << '\n';
}
return 0;
}
The program output is
0: 0
1: 1
2: 1
3: 2
4: 3
5: 5
6: 8
7: 13
8: 21
9: 34

Answer 2

int numbers[n+2]; is the declaration of an array of ints with space for n + 2 ints, this is a variable lenght array and is not part of C++ standard, though some compilers allow it it's not somenthing you should use.
If you need a variable lenght array use std::vector.
With int numbers[n+2]; if n is equal to 0 you still have space for 2 ints, if you have int numbers[n]; the array will have space for 0 ints, so the code will fail because you are trying to access memory that does not exist with numbers[0] and numbers[1].
There are several good ways to implement the Fibonacci sequence, in the site you can find many questions regarding this matter in several programming languages, here is one of them Fibonacci series in C++
Edit
So I've seen your comments about using a vector, for making the sequence you wouldn't need the vector just two variables to store the two numbers to add, to store the sequence in a vactor, you can do somenthing like:
#include <iostream>
#include <vector>
#include <iomanip>
//passing the vector by reference
void fastFibonacci(unsigned long long n, std::vector<unsigned long long>& sequence) {
unsigned long long first = 0;
unsigned long long second = 1;
sequence.push_back(first); //add first values to the vector
sequence.push_back(second); //add first values to the vector
for (unsigned long long i = 0, value = 0; i < n && value <= LLONG_MAX ; ++i) {
value = first + second;
first = second;
second = value;
sequence.push_back(value); //adding values to the vector
}
}
int main() {
unsigned long long limit; //number of values in the sequence
int num = 1;
std::vector<unsigned long long> sequence; //container for the sequence
std::cout << "Enter upper limit: ";
std::cin >> limit;
fastFibonacci(limit, sequence);
//print the sequence in a range based loop formatted with <iomanip> library
for(auto& i : sequence){
std::cout << std::setw(4) << std::left << num++ << " " << i << std::endl;
}
return 0;
}
If you want to print just one of the numbers in the sequence, just use, for instance:
std::cout << sequence[10];
Instead of the whole vector.
The code you post in the comment to the other answer won't work because the access to the vector is out of bounds in numbers[i] = numbers[i - 1] + numbers[i - 2];, if for instance i = 5, your vector only has 2 nodes but you are accessing the 6th node numbers[5].