无法迭代Arc Mutex

Question

考虑下面的代码,我将每个线程附加到Vector,以便在我生成每个线程之后将它们连接到主线程,但是我无法调用iter()JoinHandlers的向量.

我该怎么做呢？

fn main() {
    let requests = Arc::new(Mutex::new(Vec::new()));
    let threads = Arc::new(Mutex::new(Vec::new()));

    for _x in 0..100 {
        println!("Spawning thread: {}", _x);

        let mut client = Client::new();
        let thread_items = requests.clone();

        let handle = thread::spawn(move || {
            for _y in 0..100 {
                println!("Firing requests: {}", _y);

                let start = time::precise_time_s();

                let _res = client.get("http://jacob.uk.com")
                    .header(Connection::close()) 
                    .send().unwrap();

                let end = time::precise_time_s();

                thread_items.lock().unwrap().push((Request::new(end-start)));
            }
        });

        threads.lock().unwrap().push((handle));
    }

    // src/main.rs:53:22: 53:30 error: type `alloc::arc::Arc<std::sync::mutex::Mutex<collections::vec::Vec<std::thread::JoinHandle<()>>>>` does not implement any method in scope named `unwrap`
    for t in threads.iter(){
        println!("Hello World");
    }
}

Answer 1

首先,你不需要threads被包含在Arc在Mutex.你可以保持它Vec:

let mut threads = Vec::new();
...
threads.push(handle);

这是因为你没有threads在线程之间共享.您只能从主线程访问它.

其次,如果由于某种原因,你就需要保持它Arc(例如,如果你的例子并不反映你的程序,它是更为复杂的实际结构),那么你就需要锁定互斥获取到包含向量的参考,只是就像你推动时一样:

for t in threads.lock().unwrap().iter() {
    ...
}