Fibonacci Without Using Recursion

Question

How I can print the (n)th element of the Fibonacci sequence, using and without using recursion, I've solved the first half of the question [revFibo() using recursion], but it seems that I can't see the problem with my iterative answer (Fibo() Function), it keeps printing the same parameter that I gave it, but throw a trash value when I put m = 3.

#include <stdio.h>

int revfibo(int n);
int fibo(int m);

int main() {
    int n;
    printf("Give n: \n");
    scanf("%d",&n);
    printf("The Result is %d\n", fibo(n));
    return 0;
}

int revfibo(int n){
    if (n==0){
        return 0;
    } else{
        if(n==1){
            return 1;
        }else{
            return revfibo(n-1)+ revfibo(n-2);
        }
    }
}

int fibo(int m){
    int T[m+1];
    int i;
    if (m == 0){
        return 0;
    }
    else {
        if (m == 1) {
            return 1;
        } else {
            T[0] = 0;
            T[1] = 1;
            for (i = 2; i <=m; i++) {
                T[i] = T[i - 2] + T[i - 1];
            }
            return T[m];
        }
    }
}

Answer 1

Your code is correct, in that it's calculating the sequence correctly, it's just returning the wrong part of it.

The commenters are correct that it's not very elegant code, but that's not really the point of your question.

it helps to rewrite your main function as follows so you can compare values and see how the sequence develops. As you'll see your two functions produce the same result:

int main() {
    int n;
    printf("Give n: \n");
    scanf("%d",&n);

    for (int i = 3; i < n; i++)
        {
        printf("%i recursive: %i, iterative: %i\n", i, revfibo(i), fibo(i));
        }

    return 0;
}

here's an interactive compiler online you can test it with the fully modified source.

https://onlinegdb.com/P_DDaqCpw

Give n: 
10
3 recursive: 2, iterative: 2
4 recursive: 3, iterative: 3
5 recursive: 5, iterative: 5
6 recursive: 8, iterative: 8
7 recursive: 13, iterative: 13
8 recursive: 21, iterative: 21
9 recursive: 34, iterative: 34

The expected sequence is the series of numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34

Answer 2

This one is in javascript, but it meets what you are asking for:

const fibonacci = (num) => {
    if (num < 2) {
        return num;
    }
    let prev1 = 1, prev2 = 0;
    for (let i = 2; i < num ; i ++)
    {
        const tmp = prev1;
        prev1 = prev1 + prev2;
        prev2 = tmp;
    }
    return prev1 + prev2;
}

console.log(fibonacci(1)) // 1
console.log(fibonacci(2)) // 1
console.log(fibonacci(3)) // 2
console.log(fibonacci(4)) // 3
console.log(fibonacci(5)) // 5
console.log(fibonacci(6)) // 8
console.log(fibonacci(7)) // 13
console.log(fibonacci(8)) // 21
console.log(fibonacci(9)) // 34