为何忽略info参数的闭包仍能满足async_task的Schedule trait约束？

我在查看async_task库的spawn_local文档时，注意到示例代码如下：

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);

spawn_local的函数签名对Schedule trait有约束：

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,

其中Schedule trait定义如下：

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

示例中的闭包let schedule = move |runnable| s.send(runnable).unwrap();未接收info: ScheduleInfo参数，想知道它是如何满足该trait约束的。

这是因为async_task库为不同参数数量的闭包提供了Schedule trait的自动适配实现。

Schedule trait要求的schedule方法包含三个参数：&self、runnable、info，但Rust的闭包支持自动适配参数更少的场景——当你定义一个仅接收runnable的闭包时，库内部会自动生成对应的Schedule实现，在调用时直接忽略掉多余的info参数，相当于帮你补全了如下逻辑：

impl<F> Schedule for F
where
    F: Fn(Runnable),
{
    fn schedule(&self, runnable: Runnable, _info: ScheduleInfo) {
        self(runnable);
    }
}

借助这种适配，你的闭包不需要显式处理info参数，也能满足Schedule trait的约束。

内容的提问来源于stack exchange，提问作者cogle