Drop

Like the mutable lists, we have a the recursive destructor problem. Admittedly, this isn't as bad of a problem for the immutable list: if we ever hit another node that's the head of another list somewhere, we won't recursively drop it. However it's still a thing we should care about, and how to deal with isn't as clear. Here's how we solved it before:

fn main() { impl<T> Drop for List<T> { fn drop(&mut self) { let mut cur_link = self.head.take(); while let Some(mut boxed_node) = cur_link { cur_link = boxed_node.next.take(); } } } }
impl<T> Drop for List<T> {
    fn drop(&mut self) {
        let mut cur_link = self.head.take();
        while let Some(mut boxed_node) = cur_link {
            cur_link = boxed_node.next.take();
        }
    }
}

The problem is the body of the loop:

fn main() { cur_link = boxed_node.next.take(); }
cur_link = boxed_node.next.take();

This is mutating the Node inside the Box, but we can't do that with Rc; it only gives us shared access. There's two ways to handle this.

The first way is that we can keep grabbing the tail of the list and dropping the previous one to decrement its count. This will prevent the old list from recursively dropping the rest of the list because we hold an outstanding reference to it. This has the unfortunate problem that we traverse the entire list whenever we drop it. In particular this means building a list of length n in place takes O(n2) as we traverse a lists of length n-1, n-2, .., 1 to guard against overflow (this is really really really really bad).

The second way is if we could identify that we're the last list that knows about this node, we could in principle actually move the Node out of the Rc. Then we could also know when to stop: whenver we can't hoist out the Node. For reference, the unstable function is called try_unwrap.

Rc actually lets you do this... but only in nightly Rust. Honestly, I'd rather risk blowing the stack sometimes than iterate every list whenever it gets dropped. Still if you'd rather not blow the stack, here's the first (O(n)) solution:

fn main() { impl<T> Drop for List<T> { fn drop(&mut self) { // Steal the list's head let mut cur_list = self.head.take(); while let Some(node) = cur_list { // Clone the current node's next node. cur_list = node.next.clone(); // Node dropped here. If the old node had // refcount 1, then it will be dropped and freed, but it won't // be able to fully recurse and drop its child, because we // hold another Rc to it. } } } }
impl<T> Drop for List<T> {
    fn drop(&mut self) {
        // Steal the list's head
        let mut cur_list = self.head.take();
        while let Some(node) = cur_list {
            // Clone the current node's next node.
            cur_list = node.next.clone();
            // Node dropped here. If the old node had
            // refcount 1, then it will be dropped and freed, but it won't
            // be able to fully recurse and drop its child, because we
            // hold another Rc to it.
        }
    }
}

and here's the second (amortized O(1)) solution (only works on nightly):

fn main() { impl<T> Drop for List<T> { fn drop(&mut self) { let mut head = self.head.take(); while let Some(node) = head { if let Ok(mut node) = Rc::try_unwrap(node) { head = node.next.take(); } else { break; } } } } }
impl<T> Drop for List<T> {
    fn drop(&mut self) {
        let mut head = self.head.take();
        while let Some(node) = head {
            if let Ok(mut node) = Rc::try_unwrap(node) {
                head = node.next.take();
            } else {
                break;
            }
        }
    }
}