Here is a version where the nodes can be modified from outside the tree, which I presume is what was asked for.
use std::rc::Rc;
use std::cell::RefCell;
struct Node {
a : Option<Rc<RefCell<Node>>>,
b : Option<Rc<RefCell<Node>>>,
value: int
}
impl Node {
fn new(value: int) -> Rc<RefCell<Node>> {
let node = Node {
a: None,
b: None,
value: value
};
Rc::new(RefCell::new(node))
}
}
fn main() {
let first = Node::new(0);
let second = Node::new(0);
let third = Node::new(0);
first.borrow_mut().a = Some(second.clone());
second.borrow_mut().a = Some(third.clone());
second.borrow_mut().value = 1;
third.borrow_mut().value = 2;
println!("Value of second: {}", first.borrow().a.get_ref().borrow().value);
println!("Value of third: {}", first.borrow().a.get_ref().borrow().a.get_ref().borrow().value);
}
Rc is a reference counted pointer and allows a single object to have multiple owners. It doesn't allow mutation however, so a RefCell is required which allows runtime checked mutable borrowing. That's why the code uses Rc<RefCell<Node>>. The Option type is used to represent potential children with Option<Rc<RefCell<Node>>>.
Since, the Rc type auto dereferences, it's possible to directly call RefCell methods on it. These are borrow() and borrow_mut() which return a reference and mutable reference to the underlying Node. There also exist try_borrow() and try_borrow_mut() variants which cannot fail.
get_ref() is a method of the Option type which returns a reference to the underlying Rc<RefCell<Node>>. In a real peogram we would probably want to check whether the Option contains anything beforehand.
Why does the original code not work? References &T imply non-ownership, so something else would have to own the Nodes. While it would be possible to build a tree of &Node types, it wouldn't be possible to modify the Nodes outside of the tree because once borrowed, an object cannot be modified by anything else than the borrowing object.