当这个闭包似乎丢弃了一个参数时,它是如何满足特征边界的

嗨,我正在查看的文档 asnyc_task 请注意,在示例代码中:

use async_task::Runnable;
use flume::{Receiver, Sender};
use std::rc::Rc;

thread_local! {
    // A queue that holds scheduled tasks.
    static QUEUE: (Sender<Runnable>, Receiver<Runnable>) = flume::unbounded();
}

// Make a non-Send future.
let msg: Rc<str> = "Hello, world!".into();
let future = async move {
    println!("{}", msg);
};

// A function that schedules the task when it gets woken up.
let s = QUEUE.with(|(s, _)| s.clone());
let schedule = move |runnable| s.send(runnable).unwrap();

// Create a task with the future and the schedule function.
let (runnable, task) = async_task::spawn_local(future, schedule);

当涉及到的函数签名时 ::spawn_local(...) 有一个特点 Schedule

pub fn spawn_local<F, S>(future: F, schedule: S) -> (Runnable, Task<F::Output>)
where
    F: Future + 'static,
    F::Output: 'static,
    S: Schedule + Send + Sync + 'static,

哪里 日程 具有以下特征:

pub trait Schedule<M = ()>: Sealed<M> {
    // Required method
    fn schedule(&self, runnable: Runnable<M>, info: ScheduleInfo);
}

在我看来,以下代码丢失了 info: ScheduleInfo 作为一个论点,所以我想知道这种特质界限是如何得到满足的。

let schedule = move |runnable| s.send(runnable).unwrap();