rust - 可以是 `Send` 的 RefCell 替代方案？

Question

我正在设计一种可以在线程之间发送或不发送的类型，这取决于它是否是拥有的类型。
这段代码:

use std::cell::RefCell;
use std::rc::Rc;

pub enum M<'a, T> {
    Owned(),
    Ref(Rc<RefCell<&'a mut [T]>>)
}

fn main() {
    let m: M<'static, u8> = M::Owned();
    std::thread::spawn(||{
        let m = m;
    });
}

给

error[E0277]: `Rc<RefCell<&mut [_]>>` cannot be sent between threads safely
   --> src/main.rs:11:5
    |
11  |       std::thread::spawn(||{
    |  _____^^^^^^^^^^^^^^^^^^_-
    | |     |
    | |     `Rc<RefCell<&mut [_]>>` cannot be sent between threads safely
12  | |         let m = m;
13  | |     });
    | |_____- within this `[closure@src/main.rs:11:24: 13:6]`

我知道这不是 Send因为 RefCell是 !Send .我认为这是唯一阻止 M 的原因来自 Send .有没有 RefCell替代方案是 Send ?

Answer 1

I understand that this is not Send because RefCell is !Send. I think this is the only thing preventing M from being Send.

是的，这是正确的。

Is there a RefCell alternative that is Send?

当然，至少有两个。如果您只想要 &mut在另一个线程中，您实际上可以发送它:

#[derive(Debug)]
pub enum M<'a, T> {
    Owned(),
    Ref(&'a mut [T])
}

fn main() {
    let m: M<'static, u8> = M::Owned();
    std::thread::spawn(move ||{
        println!("{:?}", m);
    }).join().unwrap();
}

但是如果你使用 Rc<RefCell<_>>例如让您的 &mut可共享，您可以使用 Arc<Mutex<_>> 在多线程上下文中实现相同的功能，正如@Locke 已经提到的:

use std::sync::{Arc, Mutex};

#[derive(Debug, Clone)]
pub enum M<'a, T> {
    Owned(),
    Ref(Arc<Mutex<&'a mut [T]>>)
}

fn main() {
    let m: M<'static, u8> = M::Owned();
    let m2 = m.clone();
    std::thread::spawn(move ||{
        println!("{:?}", m2);
    }).join().unwrap();
    println!("{:?}", m);
}