异步 – 使用async(tokio)rust-websocket在客户端之间共享可变状态

2019年7月28日 638次阅读

我正在使用rust-websocket及其基于Tokio的异步系统在Rust中编写一个websocket服务器.我可以很好地为客户服务,但是,我无法弄清楚如何在客户端之间共享可变状态.以下是一些(部分)代码,演示了此问题：

let mut core = Core::new().unwrap();
let handle = core.handle();
let server = Server::bind("localhost:62831", &handle).unwrap();

let mut state = State{
    ...
};

let f = server.incoming()
    .map_err(|InvalidConnection {error, ..}| error)
    .for_each(|upgrade, _)| {
        let f = upgrade.accept()
            .and_then(|s, _| {
                let ctx = ClientContext{
                    // some other per-client values
                    state: &mut state,
                }
                ...
                return s.send(Message::binary(data).into())
                    .and_then(move |s| Ok(s, ctx)); // this could be the complete wrong way to insert context into the stream
            }).and_then(|s, ctx| {
                // client handling code here
            });

            handle.spawn(f
                .map_err(...)
                .map(...)
            );
            return Ok(())
    });

core.run(f).unwrap();

这段代码错误：

error[E0373]: closure may outlive the current function, but it borrows `**state`, which is owned by the current function
   --> src/main.rs:111:27
    |
111 |                 .and_then(|(s, _)| {
    |                           ^^^^^^^^ may outlive borrowed value `**state`
...
114 |                         state: &mut state,
    |                                     ----- `**state` is borrowed here
    |
help: to force the closure to take ownership of `**state` (and any other referenced variables), use the `move` keyword, as shown:
    |                 .and_then(move |(s, _)| {

在尝试编译器的建议时,我得到了这个：

error[E0507]: cannot move out of captured outer variable in an `FnMut` closure
   --> src/main.rs:111:27
    |
111 |                 .and_then(move |(s, _)| {
    |                           ^^^^^^^^^^^^^ cannot move out of captured outer variable in an `FnMut` closure

error: `state` does not live long enough
   --> src/main.rs:114:37
    |
114 |                         state: &mut state,
    |                                     ^^^^^ does not live long enough
...
122 |                 })
    |                 - borrowed value only lives until here
    |
    = note: borrowed value must be valid for the static lifetime...

我还尝试将状态包装在RefCell中(在状态本身之后立即创建RefCell),但是,编译器提供类似的移动错误,因为它尝试将RefCell移动到创建客户端上下文的闭包中.

最佳答案 你与
RefCell非常接近.你现在需要的是一个包装RefCell的
Rc,这样你就可以克隆Rc而不是捕获RefCell本身.

let shared_state = Rc::new(RefCell::new(State::new())));
incoming().for_each(move |s, _| {
    let shared_state = shared_state.clone();  // Left uncaptured
    shared_state.borrow_mut().do_mutable_state_stuff(); // Could panic
});

请注意,由于您现在正在使用Rc和RefCell,因此您可能需要继续将ClientContext结构转换为存储Rc>而不是一个& mut状态.对于某些事情,可能会继续使用& mut状态,但是你的& mut状态将与RefMut的生命周期联系在一起,并且如果你保持活着直到下一次关闭运行,则借用将会出现恐慌(或者失败如果你使用try_ variants).

另外请记住,如果您决定在反应堆中有多个线程,则只需将Rc更改为Arc,将RefCell更改为Mutex,这在需要时是非常自然的转换.

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