Use crate::List::{Cons, Nil}; use std::cell::RefCell; use std::rc::Rc; #[derive(Debug)] enum List {

Question

Use crate::List::{Cons, Nil}; use std::cell::RefCell; use std::rc::Rc; #[derive(Debug)] enum List {

Cons(i32, RefCell<Rc<List>>),
Nil,
}

impl List {
fn tail(&self) -> Option<&RefCell<Rc<List>>> {
match self {
Cons(_, item) => Some(item),
Nil => None,
}
}
}

fn main() {
let a = Rc::new(Cons(5, RefCell::new(Rc::new(Nil))));

println!("a initial rc count = {}", Rc::strong_count(&a));
println!("a next item = {:?}", a.tail());

let b = Rc::new(Cons(10, RefCell::new(Rc::clone(&a))));

println!("a rc count after b creation = {}", Rc::strong_count(&a));
println!("b initial rc count = {}", Rc::strong_count(&b));
println!("b next item = {:?}", b.tail());

if let Some(link) = a.tail() {
*link.borrow_mut() = Rc::clone(&b);
}

println!("b rc count after changing a = {}", Rc::strong_count(&b));
println!("a rc count after changing a = {}", Rc::strong_count(&a));

// Uncomment the next line to see that we have a cycle;
// it will overflow the stack
// println!("a next item = {:?}", a.tail());
}

#backend #embedded #programming #russian #rust

0

20.08.2020

2 ответов

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Αλεχ Zhukovsky

trait Node<T> {} #[derive(Debug)] struct Nil; impl<T> Node<T> for Nil {} #[derive(Debug)] struct LinkedList<T, U: Node<T>> { head: T, tail: U, } fn nil() -> Nil { Nil {} } fn cons<T, U: Node<T>>(head: T, tail: U) -> LinkedList<T, U> { LinkedList { head, tail } } impl<T, U: Node<T>> Node<T> for LinkedList<T, U> {} fn main() { let list = cons(10, cons(20, nil())); println!("{:?}", list); } написал такой без рц и никаких стаковерфлоу

0

20.08.2020

Αλεχ Zhukovsky · Accepted Answer

Αλεχ Zhukovsky

RefCell<Rc> назвать незаметным сложно

0

20.08.2020

172 похожих чатов

Use crate::List::{Cons, Nil}; use std::cell::RefCell; use std::rc::Rc; #[derive(Debug)] enum List {

2 ответов

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