How do I share a mutable object between threads us

2020-01-24 10:03发布

I'm trying to share a mutable object between threads in Rust using Arc, but I get this error:

error[E0596]: cannot borrow data in a `&` reference as mutable
  --> src/main.rs:11:13
   |
11 |             shared_stats_clone.add_stats();
   |             ^^^^^^^^^^^^^^^^^^ cannot borrow as mutable

This is the sample code:

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

fn main() {
    let total_stats = Stats::new();
    let shared_stats = Arc::new(total_stats);

    let threads = 5;
    for _ in 0..threads {
        let mut shared_stats_clone = shared_stats.clone();
        thread::spawn(move || {
            shared_stats_clone.add_stats();
        });
    }
}

struct Stats {
    hello: u32,
}

impl Stats {
    pub fn new() -> Stats {
        Stats { hello: 0 }
    }

    pub fn add_stats(&mut self) {
        self.hello += 1;
    }
}

What can I do?

1条回答
We Are One
2楼-- · 2020-01-24 10:15

Arc's documentation says:

Shared references in Rust disallow mutation by default, and Arc is no exception: you cannot generally obtain a mutable reference to something inside an Arc. If you need to mutate through an Arc, use Mutex, RwLock, or one of the Atomic types.

You will likely want a Mutex combined with an Arc:

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

struct Stats;

impl Stats {
    fn add_stats(&mut self, _other: &Stats) {}
}

fn main() {
    let shared_stats = Arc::new(Mutex::new(Stats));

    let threads = 5;
    for _ in 0..threads {
        let my_stats = shared_stats.clone();
        thread::spawn(move || {
            let mut shared = my_stats.lock().unwrap();
            shared.add_stats(&Stats);
        });
        // Note: Immediately joining, no multithreading happening!
        // THIS WAS A LIE, see below
    }
}

This is largely cribbed from the Mutex documentation.

How can I use shared_stats after the for? (I'm talking about the Stats object). It seems that the shared_stats cannot be easily converted to Stats.

As of Rust 1.15, it's possible to get the value back. See my additional answer for another solution as well.

[A comment in the example] says that there is no multithreading. Why?

Because I got confused! :-)

In the example code, the result of thread::spawn (a JoinHandle) is immediately dropped because it's not stored anywhere. When the handle is dropped, the thread is detached and may or may not ever finish. I was confusing it with JoinGuard, a old, removed API that joined when it is dropped. Sorry for the confusion!


For a bit of editorial, I suggest avoiding mutability completely:

use std::{ops::Add, thread};

#[derive(Debug)]
struct Stats(u64);

// Implement addition on our type
impl Add for Stats {
    type Output = Stats;
    fn add(self, other: Stats) -> Stats {
        Stats(self.0 + other.0)
    }
}

fn main() {
    let threads = 5;

    // Start threads to do computation
    let threads: Vec<_> = (0..threads).map(|_| thread::spawn(|| Stats(4))).collect();

    // Join all the threads, fail if any of them failed
    let result: Result<Vec<_>, _> = threads.into_iter().map(|t| t.join()).collect();
    let result = result.unwrap();

    // Add up all the results
    let sum = result.into_iter().fold(Stats(0), |i, sum| sum + i);
    println!("{:?}", sum);
}

Here, we keep a reference to the JoinHandle and then wait for all the threads to finish. We then collect the results and add them all up. This is the common map-reduce pattern. Note that no thread needs any mutability, it all happens in the master thread.

查看更多
登录 后发表回答