像可变列表一样,我们有一个递归析构函数问题。 诚然,对于不可变的列表来说,这并不是一个问题:如果我们碰到了另一个节点,该节点是另一个列表的头部,则不会递归地删除它。 但是,这仍然是我们应该关注的事情,如何处理尚不清楚。 这是我们之前解决问题的方法:
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();
}
}
}
cur_link = boxed_node.next.take();
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;
}
}
}
}
cargo test
Compiling lists v0.1.0 (/Users/ABeingessner/dev/too-many-lists/lists)
Finished dev [unoptimized + debuginfo] target(s) in 1.10s
Running /Users/ABeingessner/dev/too-many-lists/lists/target/debug/deps/lists-86544f1d97438f1f
running 8 tests
test first::test::basics ... ok
test second::test::basics ... ok
test second::test::into_iter ... ok
test second::test::iter ... ok
test second::test::iter_mut ... ok
test second::test::peek ... ok
test third::test::basics ... ok
test third::test::iter ... ok
test result: ok. 8 passed; 0 failed; 0 ignored; 0 measured; 0 filtered out