线程引用需要静态生存期吗? [英] Thread references require static lifetime?

问题描述

从直觉上讲，传递给派生线程的引用需要具有静态生存期，但我不清楚究竟是什么导致以下代码无法编译:

While it makes sense intuitively that references passed to spawned threads need to have static lifetimes, I'm unclear about what exactly is making the following code not compile:

use std::sync::Arc;
use std::sync::Mutex;

struct M;

fn do_something(m : Arc<Mutex<&M>>) {
    println!("Ha, do nothing!");
}

fn main() {
    let a = M;
    {
        let c : Arc<Mutex<&M>> = Arc::new(Mutex::new(&a));
        for i in 0..2 {
            let c_clone = c.clone();
            ::std::thread::spawn(move || do_something(c_clone));
        }
    }
}

编译此小程序会出现以下错误:

Compiling this small program gives the following error:

$ rustc -o test test.rs
test.rs:13:55: 13:56 error: `a` does not live long enough
test.rs:13         let c : Arc<Mutex<&M>> = Arc::new(Mutex::new(&a));
                                                             ^
note: reference must be valid for the static lifetime...

在我看来，变量a的寿命将超过c_clone，在这种情况下，这很重要...?希望有人可以帮助我了解我的缺失！

It seems to me that the variable a will out-live c_clone, which is what matters in this case...? Hopefully someone can help me understand what I'm missing!

推荐答案

本质上，Arc和Mutex包装是多余的:您正在传递对本地堆栈上的内容的引用.当您使用std::thread::spawn生成线程时，没有任何将生存期链接在一起的内容.主线程可以自由地结束并释放其中的任何内容(在这种情况下，包括a在内)，甚至可以在其他线程生成之前甚至执行 start 来释放它.因此，在这种情况下，a可以在生成的线程执行任何操作时引用释放的内存，而将c_clone保留为悬空指针.这就是为什么生成的线程关闭的环境必须为'static

In its essence, the Arc and Mutex wrapping is superfluous: you are passing a reference to something on the local stack. When you spawn a thread with std::thread::spawn, there is nothing linking the lifetimes together; the main thread is quite at liberty to conclude and free anything in it—in this case, including a—before any other threads it spawns even start executing; thus in this case a could refer to freed memory by the time the spawned thread does anything, leaving c_clone as a dangling pointer. This is why the environment of the closure of a spawned thread must be 'static.

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