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我正在使用 rust-websocket 及其基于 Tokio 的异步系​​统在 Rust 中编写 websocket 服务器。我可以很好地为客户服务，但是，我不知道如何在客户之间共享可变状态。这是一些演示此问题的（部分）代码：

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();

此代码错误：

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 a 中RefCell（在状态本身之后创建RefCell右侧），但是，编译器给出了类似的移动错误，因为它试图将 a 移动RefCell到创建客户端上下文的闭包中。

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1 回答 1

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你与RefCell. 你现在需要的是一个Rc包装，RefCell这样你就可以克隆Rc而不是捕获它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，由于您现在使用的是Rc's 和RefCell's，因此您可能需要继续将ClientContext结构转换为存储 Rc> 而不是&mut State. &mut State在某些事情上可能会继续使用's，但你的 ' s&mut State将RefMut与try_）。

另外请记住，如果您决定要在反应器中拥有多个线程，则只需更改Rc为Arc和RefCell，Mutex这是在需要时非常自然的转换。

于 2017-07-03T17:50:21.527 回答