如何使用没有作用域的RwLocks？

Question

我试图在不使用范围线程的情况下在多个线程中共享一个RwLock,但我无法弄清楚如何使生命周期正确.我认为这是可能的(否则RwLocks的重点是什么？)但是我找不到它的任何例子.

这是我想要完成的一个玩具示例.任何意见,将不胜感激. 这个代码的防锈围栏

use std::sync::{Arc, RwLock};
use std::thread;

struct Stuff {
    x: i32
}

fn main() {
    let mut stuff = Stuff{x: 5};
    helper(&mut stuff);
    println!("done");
}

fn helper(stuff: &mut Stuff){
    let rwlock = RwLock::new(stuff);
    let arc = Arc::new(rwlock);
    let local_arc = arc.clone();
    for _ in 0..10{
        let my_rwlock = arc.clone();
        thread::spawn(move || {
            let reader = my_rwlock.read().unwrap();
            // do some stuff
        });
    }
    let mut writer = local_arc.write().unwrap();
    writer.x += 1;
}

Answer 1

&mut发送到非作用域线程的引用是不安全的,因为在引用的数据被释放后,该线程仍然可以运行.此外,在helper返回之后,主线程仍然可以变异stuff,并且生成的线程也可以stuff间接地变异,这在Rust中是不允许的(变量只能有一个可变别名).

相反,它RwLock应该拥有数据,而不是借用它.这意味着helper应该得到一个Stuff而不是一个&mut Stuff.

use std::sync::{Arc, RwLock};
use std::thread;

struct Stuff {
    x: i32
}

fn main() {
    let mut stuff = Stuff{x: 5};
    helper(stuff);
    println!("done");
}

fn helper(stuff: Stuff){
    let rwlock = RwLock::new(stuff);
    let arc = Arc::new(rwlock);
    let local_arc = arc.clone();
    for _ in 0..10{
        let my_rwlock = arc.clone();
        thread::spawn(move || {
            let reader = my_rwlock.read().unwrap();
            // do some stuff
        });
    }
    let mut writer = local_arc.write().unwrap();
    writer.x += 1;
}