Why does mutating a passed Option inside a function not propagate to the Option outside?

Question

I've illustrated it thus:

fn main() {

    let mut opt1 = Some(1);
    // compiler complains that opt2 doesn't have to be mutable
    let mut opt2 = Some(1);

    fn take_inner(mut opt: Option<u8>) {
        opt.take();
    };

    opt1.take();
    take_inner(opt2);

    println!("opt1 {:?}", opt1); // prints "opt1 None"
    println!("opt2 {:?}", opt2); // prints "opt2 Some(1)"

}

Rust Playground link

Why does calling opt.take() within a function have a different effect than calling it outside (relative to the main function's scope)?

Answer 1

Since u8 is a copy type, and Option has

impl<T> Copy for Option<T>
where
    T: Copy,

take_inner takes a copy of opt2

You can fix that by taking a mutable reference:

fn main() {
    let mut opt1 = Some(1);
    // compiler complains that opt2 doesn't have to be mutable
    let mut opt2 = Some(1);

    fn take_inner(opt: &mut Option<u8>) {
        opt.take();
    };

    opt1.take();
    take_inner(&mut opt2);

    println!("opt1 {:?}", opt1);
    println!("opt2 {:?}", opt2);
}

Playground link