闭包内的 Arc<Mutex<T>>：期望实现 `Fn` 特征的闭包，但此闭包仅实现 `FnOnce`

Question

我正在尝试与闭包共享Arc<Mutex<T>>。我收到closure is FnOncebecause it moves the variableshared_wake_deque_ out of its environment的错误：

use core::task::{Context, Poll, Waker};
use hyper::client::connect::{Connection, Connected};
use std::sync::{Arc, Mutex};
use std::collections::VecDeque;

pub type OnPollRead = Arc<dyn Fn(&mut Context<'_>, &mut tokio::io::ReadBuf<'_>) -> Poll<std::io::Result<()>> + Send + Sync>;
pub type OnPollWrite = Arc<dyn Fn(&mut Context<'_>, &[u8]) -> Poll<core::result::Result<usize, std::io::Error>> + Send + Sync>;

#[derive(Clone)]
pub struct CustomTransporter {
    on_poll_read: Option<OnPollRead>,
    on_poll_write: Option<OnPollWrite>,
}

pub struct OVPNClient {
}

impl OVPNClient {
    pub fn send(&self, buffer: &[u8]) {
        
    }
}

unsafe impl Send for OVPNClient {}


unsafe impl Send for CustomTransporter {}

impl CustomTransporter {
    pub fn new(on_poll_read: Option<OnPollRead>, on_poll_write: Option<OnPollWrite>) -> CustomTransporter {
        CustomTransporter{
            on_poll_read: on_poll_read,
            on_poll_write: on_poll_write
        }
    }
}

fn main () {
    let openvpn_client = Arc::new(Mutex::new(OVPNClient{}));

    let shared_wake_deque = Arc::new(Mutex::new(VecDeque::<Waker>::new()));
    
    let shared_wake_deque_ = shared_wake_deque.clone();
    let on_poll_read = Arc::new(|context: &mut Context, buffer: &mut tokio::io::ReadBuf| -> Poll<std::io::Result<()>> {
        let shared_wake_deque__ = shared_wake_deque_;
        Poll::Ready(Ok(()))
    });

    let on_poll_write = Arc::new(|context: &mut Context, buffer: &[u8]| -> Poll<core::result::Result<usize, std::io::Error>>{
        openvpn_client.lock().unwrap().send(buffer);
        Poll::Ready(Ok(0))
    });

    let connector = CustomTransporter::new(Some(on_poll_read), Some(on_poll_write));
}

错误：

error[E0525]: expected a closure that implements the `Fn` trait, but this closure only implements `FnOnce`
  --> src/main.rs:44:33
   |
44 |     let on_poll_read = Arc::new(|context: &mut Context, buffer: &mut tokio::io::ReadBuf| -> Poll<std::io::Result<()>> {
   |                                 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ this closure implements `FnOnce`, not `Fn`
45 |         let shared_wake_deque__ = shared_wake_deque_;
   |                                   ------------------ closure is `FnOnce` because it moves the variable `shared_wake_deque_` out of its environment
...
54 |     let connector = CustomTransporter::new(Some(on_poll_read), Some(on_poll_write));
   |                                                 ------------ the requirement to implement `Fn` derives from here

我知道移动只能发生一次。我什至没有关闭move。如您所见，我正在尝试使用Arc<Mutex<T>> 的克隆，所以我并没有完全移动它。你可以看到我尝试了克隆，所以我移动了一个仅在此闭包上使用的变量，并且不再使用，但它不起作用。

https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=d996163f5d2a6e9322af9a7472581ce6

Answer 1

因为你有

let shared_wake_deque__ = shared_wake_deque_;

作为一个简单的赋值，和

没有区别

let foo = vec![];
let foo2 = foo;

运行代码将 vec 的所有权从 foo 转移到 foo2。因为您在 lambda 中具有此功能，所以运行显式将 shared_wake_deque_ 值的所有权转移到 shared_wake_deque__，这意味着不能再使用 shared_wake_deque_。为此，要求on_poll_read 只能运行一次，否则会有多个事物试图拥有同一个对象（两者都是on_poll_read，因为它需要拥有它才能在一些未来的电话，而之前任何电话中的shared_wake_deque__ 也仍然必须拥有它）。由于这将违反 Rust 的安全保证，因为某些东西只能归一物所有，因此您会收到此错误。

很可能，您想要的是给我们一个参考，例如

let shared_wake_deque__ = &shared_wake_deque_;

通过使用对shared_wake_deque_ 的引用，允许闭包多次运行并给出多个共享引用。但是，仅凭这一点，您就会得到

error[E0597]: `shared_wake_deque_` does not live long enough

因为你试图引用一个在闭包外声明的变量，而闭包可以在shared_wake_deque_ 被删除后运行。为了解决那个，你需要闭包来获得shared_wake_deque_的所有权，这是通过使用move关键字来完成的，例如

let on_poll_read = Arc::new(move |context: &mut Context, buffer: &mut tokio::io::ReadBuf| -> Poll<std::io::Result<()>> {
    let shared_wake_deque__ = &shared_wake_deque_;

那么on_poll_write 也会出现同样的问题，所以也需要move：

let on_poll_write = Arc::new(move |context: &mut Context, buffer: &[u8]| -> Poll<core::result::Result<usize, std::io::Error>>{

(Rust playground)