我正在尝试在Rust中创建一个节点网络,我希望网络中的每个节点都能够知道每个其他连接节点。我认为这可以通过弱Rc
来完成,如下所示:
use std::cell::Cell;
use std::cell::RefCell;
use std::rc::Rc;
use std::rc::Weak;
struct Node {
name: String,
known_nodes: Rc<RefCell<Vec<Weak<Node>>>>,
}
impl Node {
fn connect_to_network(&mut self) {
self.known_nodes
.borrow_mut()
.push(Rc::downgrade(&Rc::new(*self)));
}
}
fn main() {
let known_nodes = Rc::new(RefCell::new(Vec::new()));
let node_one = Node {
name: "node1",
known_nodes: known_nodes.copy(),
};
node_one.connect_to_network();
let node_two = Node {
name: "node2",
known_nodes: known_nodes.copy(),
};
node_two.connect_to_network();
}
然而这会产生
无法摆脱借来的内容
在:
self.known_senders.borrow_mut().push(Rc::downgrade(&Rc::new(*self)));
因为*self
已移出&Rc::new(*self)
中的借用内容。
关于每个节点如何跟踪网络中所有其他节点的任何想法?
答案 0 :(得分:3)
您应该将节点和网络分开,因为您的网络必须获取节点的所有权才能创建Rc
(或者至少,它必须采用已创建的Rc
1}})。这是一个更好的设计,可以达到你想要的效果:
use std::rc::Rc;
use std::rc::Weak;
use std::cell::RefCell;
#[derive(Debug)]
struct Node {
name: String,
}
#[derive(Default, Debug)]
struct Network {
nodes: Rc<RefCell<Vec<Weak<Node>>>>,
}
impl Network {
fn add_node(&mut self, node: Node) -> Rc<Node> {
let node = Rc::new(node);
self.nodes.borrow_mut().push(Rc::downgrade(&node));
node
}
}
fn main() {
let mut network = Network::default();
let node_1 = Node { name: "node_1".into() };
let node_2 = Node { name: "node_2".into() };
let _node_1 = network.add_node(node_1);
let _node_2 = network.add_node(node_2);
}
如果要存储对self
的引用,可以执行以下操作:
use std::cell::RefCell;
use std::rc::Rc;
use std::rc::Weak;
type MutableNode = Rc<RefCell<Node>>;
type Network = Rc<RefCell<Vec<Weak<RefCell<Node>>>>>;
struct Node {
name: String,
others: Network,
}
impl Node {
fn new(name: String) -> MutableNode {
let node = Rc::new(RefCell::new(Node {
name,
others: Rc::new(RefCell::new(Vec::new())),
}));
{
let tmp = node.borrow();
tmp.others.borrow_mut().push(Rc::downgrade(&node));
}
node
}
fn add_node(&mut self, name: String) -> MutableNode {
let others = self.others.clone();
let node = Rc::new(RefCell::new(Node { name, others }));
self.others
.borrow_mut()
.push(Rc::downgrade(&node));
node
}
fn len(&self) -> usize {
self.others.borrow().len()
}
}
fn main() {
let node_0 = Node::new("node_0".into());
let node_1 = node_0.borrow_mut().add_node("node_1".into());
let node_2 = node_0.borrow_mut().add_node("node_2".into());
assert_eq!(node_0.borrow().len(), 3);
assert_eq!(node_1.borrow().len(), 3);
assert_eq!(node_2.borrow().len(), 3);
}
答案 1 :(得分:0)
Rc::new(value:T)
使用value
。您的函数只借用它,因此您无法调用Rc::new(*self)
我建议你像上面的答案一样创建一个网络结构。或者您可以将节点包装在Rc<RefCell<Node>>
中,如下所示:
use std::cell::RefCell;
use std::rc::Rc;
use std::rc::Weak;
#[derive(Debug)]
struct Node {
name: String,
known_nodes: Rc<RefCell<Vec<Weak<RefCell<Node>>>>>,
}
impl Node {
fn connect_to_network(&mut self,ref_to_self: Weak<RefCell<Node>>) {
self.known_nodes
.borrow_mut()
.push(ref_to_self);
}
}
fn main() {
let known_nodes = Rc::new(RefCell::new(Vec::new()));
let node_one = Rc::new(RefCell::new(Node {
name: "node1".into(),
known_nodes: known_nodes.clone(),
}));
node_one.borrow_mut().connect_to_network(Rc::downgrade(&node_one));
let node_two = Rc::new(RefCell::new(Node {
name: "node2".into(),
known_nodes: known_nodes.clone(),
}));
node_two.borrow_mut().connect_to_network(Rc::downgrade(&node_two));
println!("{:?}",known_nodes.borrow()[0].upgrade());
println!("{:?}",known_nodes.borrow()[1].upgrade());
drop(node_one);
drop(node_two);
println!("{:?}",known_nodes.borrow()[0].upgrade());
println!("{:?}",known_nodes.borrow()[1].upgrade());
}
在这种情况下,您实际上并不需要connect_to_network
功能,您可以直接将Weak<RefCell<Node>>
添加到known_nodes
如果您希望代码看起来更干净,可以像这样向Rc<RefCell<Node>>
引入新类型别名
struct Node {
name: String,
known_nodes: Rc<RefCell<Vec<Weak<RefCell<Node>>>>>,
}
type RcNode = Rc<RefCell<Node>>;
trait Connectable {
fn connect_to_network(&self);
}
impl Connectable for RcNode {
fn connect_to_network(&self){
let node = self.borrow_mut();
node.known_nodes.borrow_mut().push(Rc::downgrade(self));
}
}
所以你可以打电话给
let node_one:RcNode = Rc::new(RefCell::new(Node {
name: "node1".into(),
known_nodes: known_nodes.clone(),
}));
node_one.connect_to_network();