clone - 省略传递给函数的双克隆值

Question

我正在用 Rust 编写一个文件观察器，我遇到了双重克隆 INotify 以在 spawned 范围内使用它的问题:

fn watch_file(watcher: INotify, path: &String) {
    let mut watcher = watcher.clone();
    let path = Path::new(path);
    let pusher = RedisPusher::new("redis://127.0.0.1/0", "arkona").unwrap();

    info!("Watching `{}`", path.as_str().unwrap());

    spawn(move || {
        let _ = watcher.watch(path, |_, file| {
            info!("Change in `{}`", file.path().as_str().unwrap());
            file.read_to_string().map(|content| pusher.push(content)).ok();
        });
    });
}

fn main() {
    let args = os::args();
    let watcher = INotify::init().unwrap();

    for path in args.tail().iter() {
        watch_file(watcher.clone(), path);
    }
}

我仍在学习 Rust，我找不到任何方法将可变观察器作为值传递给此方法。这是唯一(且惯用)的方式还是有可能让它最多发生一次？

Answer 1

附带说明一下，如果您获得了参数的所有权(即参数不是引用)，则您没有在函数签名中声明可变性:

fn add_thing(things: Vec<uint>) -> Vec<uint> {
    let mut things = things; // We own this, so we can make it mutable
    things.push(1);
    things
}

fn main() {
    let things = vec![3, 2];
    let changed = add_thing(things);
    println!("{}", changed);
}