Logic about return in rust

Question

It's easy rust code, which is from rustbook on rust website. In rust if line is without ";" at the end it works like a return.

1. In this case line break counter * 2; has this and return value, why?

2. If I drop this ";" in thia line i get the same result, why?

3. Even if I add this ";" to the next line i mean }; i get the same result, why?

I try to understand it, but can't catch this logic.

fn main() {
    let mut counter = 0;

    let result = loop {
        counter += 1;

        if counter == 10 {
            break counter * 2;
        }
    };

    println!("Zmienna result wynosi {result}");
}

Answer 1

break EXPRESSION, when reached, immediately terminates the closest loop¹ (that is moves the program flow to just after the loops closing brace }) and the whole loop {…} evaluates to the value of EXPRESSION (always counter * 2 = 10 * 2 = 20 in your example), what comes after it does not matter at all.

---

^{1: or for or while but they don't allow breaking with expressions so that'd result in a compiler error}