Remove a child and clean up a recursive structure

To take an owned child out of a mutable class, leave a replacement behind. replace(node.next, Nothing) returns the old tail without cloning or allocating:

class Link:
    value: i64
    next: Option[Box[Link]]

def main() -> Result[(), AllocError | IoError]:
    mut chain = Some(Box(Link(1, Some(Box(Link(2, Nothing))?)))?)
    while True:
        match chain:
            case Some(link):
                mut node = link
                print(node.value)?
                chain = replace(node.next, Nothing)
            case Nothing:
                break
    Ok(())

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Replacing node.next with Nothing leaves that field empty when node drops. Use another tail as the replacement to relink nodes. replace requires a mutable field and evaluates the replacement before any destination indices.

Automatic cleanup can drop a deep boxed chain without growing the native stack. A recursive traversal you write can still overflow it; use a loop for unbounded depth.

To inspect a structure while keeping it, traverse it through a borrow. Moving individual owned fields out without replacing them remains unsupported. Automatic copy, equality, print, and str.repr also reject recursive data, including containers holding it. Print selected scalar fields as above, or write an explicit traversal.