CODEWITHSUNDEEP
rust
The Mungler
The Mungler

Reputation: 166

Value is still wrapped up in an Option even after if let statement

I have a struct with two fields, state and children. Children is an option that contains a vector of children. I am using if let syntax to deconstruct this option back to the vector.

fn main() {
    let mut root = Node::new2d(3);
    loop {
        root.calc_scores(100);
        if let Some(mut children) = root.children {
            let child = children.pop();
            root = Node {
                state: child.0,
                children: None,
            }
        }
        root.make_children();
    }
}
struct Node<T> {
    state: T,
    children: Option<Vec<(T, i32)>>,
}

The above code doesn't compile. It complains that child is of type Option<(Board2d,i32)>. Why is child still wrapped up in an option enum? Doesn't if let Some(mut children) = root.children{} take the vector out of the Option enum?

Upvotes: 0

Views: 323

Answers (1)

KillianDS
KillianDS

Reputation: 17186

The pop method on a Vec returns an Option, because the Vec can actually be empty and thus return None, you should check that too.

That being said, Option<Vec<_>> is not something you will often need. It only makes sense if you really want a difference between an empty vector and no vector at all, which doesn't seem to be the case here. so you'd end up with something as follows:

fn main() {
    let mut root = Node::new2d(3);
    loop {
        root.calc_scores(100);
        if let Some(child) = root.children.pop() {
            root = Node {
                state: child.0,
                children: vec![],
            }
        }
        root.make_children();
    }
}
struct Node<T> {
    state: T,
    children: Vec<(T, i32)>,
}

Upvotes: 5

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