CODEWITHSUNDEEP
rustfibonacci
Farhan Mushtaque
Farhan Mushtaque

Reputation: 11

Trying to get the nth Fibonacci number but always gives me 2^nth

I tried to solve Fibonacci series problem in rust, but unfortunately it always return me exponent of 2^n. Here is my code:

use std::io;
// Fibonacci Series Number
// F(n) = 1,1,2,3,5,8,13,....(n-1),((n-1)+n)
// where, n >= 0

fn fibo(mut n: u32) -> u32 {
    let mut _prev: u32 = 0;
    let mut current: u32 = 1;
    let mut next: u32 = 1;
    let mut sum: u32 = 1;
    if n <= 0 {
        return 0;
    }
    if n == 1 || n == 2 {
        return 1;
    };
    if n > 2 {
        while n >= 1 {
            current = next;
            _prev = current;
            sum = _prev + current;
            next = sum;
            n -= 1;
        }
        next
    } else {
        return 1;
    }
}

fn main() {
    println!("Fibo of nth: What is nth?: n --> ");
    let mut n = String::new();
    io::stdin().read_line(&mut n).expect("Invalid input");
    let n: u32 = n
        .trim()
        .parse()
        .expect("Not a number. Please type a number");
    let fibo = fibo(n);
    println!("Fibo of {n}: {fibo}");
}

Here is the output:

I tried to return next which will be the answer of the nth Fibonacci, but it giving me 2^n every time.

Upvotes: 0

Views: 72

Answers (1)

orlp
orlp

Reputation: 117771

Your update is as follows:

current = next;
_prev = current;
sum = _prev + current;
next = sum;

These don't happen all at the same time, but step by step. So after your first step current = next, current is overwritten. Then when you do _prev = current... _prev is also equal to next. Which means sum = _prev + current is just sum = 2*next. Then next = sum is essentially just equivalent to next = 2*next, which is why you're seeing powers of two.

There is a much simpler update you can do, that also only needs 2 mutable variables, current and next:

let sum = current + next;
current = next;
next = sum;

Upvotes: 3

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