如何从Arc <Mutex <T >>取得T的所有权？

Question

我想从一个受a保护的函数返回一个值Mutex,但是无法理解如何正确地执行它.此代码不起作用:

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

fn func() -> Result<(), String> {
    let result_my = Arc::new(Mutex::new(Ok(())));
    let result_his = result_my.clone();

    let t = std::thread::spawn(move || {
        let mut result = result_his.lock().unwrap();
        *result = Err("something failed".to_string());
    });

    t.join().expect("Unable to join thread");

    let guard = result_my.lock().unwrap();
    *guard
}

fn main() {
    println!("func() -> {:?}", func());
}

操场

编译器抱怨:

error[E0507]: cannot move out of borrowed content
  --> src/main.rs:16:5
   |
16 |     *guard
   |     ^^^^^^ cannot move out of borrowed content

Answer 1

在Rust 1.15中,您可以使用Arc::try_unwrap和Mutex::into_inner:

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

fn func() -> Result<(), String> {
    let result_my = Arc::new(Mutex::new(Ok(())));
    let result_thread = result_my.clone();

    let t = std::thread::spawn(move || {
        let mut result = result_thread.lock().unwrap();
        *result = Err("something failed".to_string());
    });

    t.join().expect("Unable to join threads");

    let lock = Arc::try_unwrap(result_my).expect("Lock still has multiple owners");
    lock.into_inner().expect("Mutex cannot be locked")
}

fn main() {
    println!("func() -> {:?}", func());
}

Answer 2

到目前为止我找到的最佳解决方案是将结果包装成一个Option然后将其取出:

fn func() -> Result<(), String> {
    let result_my = Arc::new(Mutex::new(Some(Ok(()))));
    let result_his = result_my.clone();

    let t = std::thread::spawn(move || {
        let mut result = result_his.lock().unwrap();
        *result = Some(Err("something failed".to_string()));
    });

    t.join().expect("Unable to join thread");

    let mut guard = result_my.lock().unwrap();
    guard.take().unwrap()
}

它似乎比mem::replace@SBSTP提出的解决方案更好,因为不需要T为交换构造一个空,它可以防止多次提取.