excursion-000

Where we visit Fibonacci in Rust style

Setup

The following needs to be prepared

• A project named fibonacci
• Classic fibonacci code

fn main() {
    println!("fibonacci({}) = {}", 0, fibonacci(0));
    println!("fibonacci({}) = {}", 1, fibonacci(1));
    println!("fibonacci({}) = {}", 2, fibonacci(2));
    println!("fibonacci({}) = {}", 3, fibonacci(3));
    println!("fibonacci({}) = {}", 4, fibonacci(4));
    println!("fibonacci({}) = {}", 5, fibonacci(5));
    println!("fibonacci({}) = {}", 6, fibonacci(6));
    println!("fibonacci({}) = {}", 7, fibonacci(7));
}

fn fibonacci(n: i32) -> i32 {
	if n == 0 {
		return 0;
	} else if n == 1 {
		return 1;
	} else {
		return fibonacci(n - 1) + fibonacci(n - 2);
	}
}

Script

Rust has some unique features. One way of getting to know these features is to explore familiar territory in a new light. In this excursion we explore a function to calculate the Fibonacci numbers. The Fibonacci numbers is a sequence starting with the numbers 0, 1, 1, 2, 3 where each new number is the sum of the preceding two numbers.

We will use the following naive implementation as a starting point to exploring Rust.

Rust is an expression based language. This means that almost every syntactic structure returns a value. In this example we can forgo with all of the return keywords.

fn fibonacci(n: i32) -> i32 {
	if n == 0 {
		0
	} else if n == 1 {
		1
	} else {
		fibonacci(n - 1) + fibonacci(n - 2)
	}
}

Note that we also have to remove the semi-colon, otherwise Rust interprets this as an empty expression returning unit, which Rust catches as an incompatible types error

fibonacci(n - 1) + fibonacci(n - 2);

The convoluted if-else-if-else construction is at least not very readably or Rust-like. Better would be to use Rusts match keyword.

	match n {
		0 => 0,
		1 => 1,
		_ => fibonacci(n - 1) + fibonacci(n - 2)
	}

We could even collapse the two initial arms using a pattern.

0...1 => n,

And there you have it, Rust-like fibonacci numbers