【发布时间】:2015-07-06 05:40:32
【问题描述】:
给定一个像这样的简单结构:
struct Server {
clients: HashMap<usize, Client>
}
以&mut 身份访问Client 的最佳方式是什么?考虑以下代码:
use std::collections::HashMap;
struct Client {
pub poked: bool
}
impl Client {
pub fn poked(&self) -> bool {
self.poked
}
pub fn set_poked(&mut self) {
self.poked = true;
}
}
struct Server {
clients: HashMap<usize, Client>
}
impl Server {
pub fn poke_client(&mut self, token: usize) {
let client = self.clients.get_mut(&token).unwrap();
self.poke(client);
}
fn poke(&self, c: &mut Client) {
c.set_poked();
}
}
fn main() {
let mut s = Server { clients: HashMap::new() };
s.clients.insert(1, Client { poked: false });
s.poke_client(1);
assert!(s.clients.get(&1).unwrap().poked() == true);
}
我看到的仅有的两个选项是在客户端中使用RefCell/Cell,这让事情看起来非常糟糕:
pub struct Client {
nickname: RefCell<Option<String>>,
username: RefCell<Option<String>>,
realname: RefCell<Option<String>>,
hostname: RefCell<Option<String>>,
out_socket: RefCell<Box<Write>>,
}
或者将clients 包裹在RefCell 中,这使得Server 不可能有像这样的简单方法:
pub fn client_by_token(&self, token: usize) -> Option<&Client> {
self.clients_tok.get(&token)
}
强迫我使用闭包(例如with_client_by_token(|c| ...))。
【问题讨论】:
-
Rust 风格指南是 4 空格缩进。
标签: hashmap rust borrow-checker