Question

我写了一个简单的方法来查看按顺序返回Tokio的DelayQueue中的项目。 DelayQueue具有方法poll_expired，该方法以task::Context作为参数。我想在循环中调用poll_explained，直到Delays中的所有DelayQueue实例都超时并由poll_expired返回并使用Tokio提供的Context运行。为此，我使用一个函数poll_fn：

use std::vec::Vec;
use tokio::time::delay_queue::Key;
use tokio::time::{DelayQueue, Duration, Error};

use std::future::Future;
use std::pin::Pin;
use std::task::{Context, Poll};

/// Creates a new future wrapping around a function returning [`Poll`].
fn poll_fn<T, F>(f: F) -> PollFn<F>
where
    F: FnMut(&mut Context<'_>) -> Poll<T>,
{
    PollFn { f }
}

PollFn实现Future的地方：

/// Future for the [`poll_fn`] function.
struct PollFn<F> {
    f: F,
}

impl<T, F> Future for PollFn<F>
where
    F: FnMut(&mut Context<'_>) -> Poll<T>,
{
    type Output = T;

    fn poll(mut self: Pin<&mut Self>, cx: &mut Context<'_>) -> Poll<T> {
        (&mut self.f)(cx)
    }
}

然后，我创建我的函数，该函数将所有到期的Delays批处理并将其委托给poll_fn：

macro_rules! ready {
    ($e:expr $(,)?) => {
        match $e {
            std::task::Poll::Ready(t) => t,
            std::task::Poll::Pending => return std::task::Poll::Pending,
        }
    };
}

fn poll_expired_batch(
    dq: &mut DelayQueue<u32>,
    cx: &mut Context<'_>,
) -> Poll<Result<Vec<(u32, u64)>, Error>> {
    let mut expired_keys: Vec<(u32, u64)> = Vec::new();
    while let Some(res) = ready!(dq.poll_expired(cx)) {
        let exp = res?;
        let data = exp.get_ref();
        expired_keys.push((data.inner, data.when));
    }
    Poll::Ready(Ok(expired_keys));
}

async fn poll_batch(dq: &mut DelayQueue<u32>) -> Result<Vec<(u32, u64)>, Error> {
    poll_fn(|cx| poll_expired_batch(dq, cx)).await;
}

然后在tokio::main块中，我有以下代码：

#[tokio::main]
async fn main() {
    let mut dq: DelayQueue<u32> = DelayQueue::new();
    let mut keys: Vec<Key> = Vec::new();
    (0..10).map(|i| keys.push(dq.insert(i, Duration::new(1, 0))));
    println!("Insertion order");
    for key in keys {
        println("Key {} inserted");
    }
    let keys_returned = (poll_batch(&mut dq).await)?;
}

我收到以下错误，但我无法找出错误的含义和解决方法：

error[E0271]: type mismatch resolving `<impl core::future::future::Future as core::future::future::Future>::Output == std::result::Result<std::vec::Vec<(u32, u64)>, tokio::time::error::Error>`
  --> examples/delay_q.rs:75:51
   |
75 | async fn poll_batch(dq : &mut DelayQueue<u32>) -> Result<Vec<(u32, u64)>, Error>{
   |                                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected (), found enum `std::result::Result`
   |
   = note: expected type `()`
              found type `std::result::Result<std::vec::Vec<(u32, u64)>, tokio::time::error::Error>`
   = note: the return type of a function must have a statically known size

异步fn中的返回类型错误：期望（），在枚举std :: result :: Result中找到

0 个答案: