我正在尝试实现一个异步读取包装器,它将添加读取超时功能。目标是API是普通的AsyncRead。换句话说,我不想在代码中添加io.read(buf).timeout(t)。相反,读取实例本身应该在给定的超时过期后返回适当的io::ErrorKind::TimedOut。
不过,我无法投票让delay做好准备。总是悬而未决的。我试过async-std,futures,smol-timeout --同样的结果。虽然超时在等待时会触发,但在民意测验中却不会触发。我知道暂停并不容易。需要一些东西来唤醒它。我做错了什么?怎么才能挺过去?
use async_std::{
future::Future,
io,
pin::Pin,
task::{sleep, Context, Poll},
};
use std::time::Duration;
pub struct PrudentIo<IO> {
expired: Option<Pin<Box<dyn Future<Output = ()> + Sync + Send>>>,
timeout: Duration,
io: IO,
}
impl<IO> PrudentIo<IO> {
pub fn new(timeout: Duration, io: IO) -> Self {
PrudentIo {
expired: None,
timeout,
io,
}
}
}
fn delay(t: Duration) -> Option<Pin<Box<dyn Future<Output = ()> + Sync + Send + 'static>>> {
if t.is_zero() {
return None;
}
Some(Box::pin(sleep(t)))
}
impl<IO: io::Read + Unpin> io::Read for PrudentIo<IO> {
fn poll_read(
mut self: Pin<&mut Self>,
cx: &mut Context<'_>,
buf: &mut [u8],
) -> Poll<io::Result<usize>> {
if let Some(ref mut expired) = self.expired {
match expired.as_mut().poll(cx) {
Poll::Ready(_) => {
println!("expired ready");
// too much time passed since last read/write
return Poll::Ready(Err(io::ErrorKind::TimedOut.into()));
}
Poll::Pending => {
println!("expired pending");
// in good time
}
}
}
let res = Pin::new(&mut self.io).poll_read(cx, buf);
println!("read {:?}", res);
match res {
Poll::Pending => {
if self.expired.is_none() {
// No data, start checking for a timeout
self.expired = delay(self.timeout);
}
}
Poll::Ready(_) => self.expired = None,
}
res
}
}
impl<IO: io::Write + Unpin> io::Write for PrudentIo<IO> {
fn poll_write(
mut self: Pin<&mut Self>,
cx: &mut Context<'_>,
buf: &[u8],
) -> Poll<io::Result<usize>> {
Pin::new(&mut self.io).poll_write(cx, buf)
}
fn poll_flush(mut self: Pin<&mut Self>, cx: &mut Context<'_>) -> Poll<io::Result<()>> {
Pin::new(&mut self.io).poll_flush(cx)
}
fn poll_close(mut self: Pin<&mut Self>, cx: &mut Context<'_>) -> Poll<io::Result<()>> {
Pin::new(&mut self.io).poll_close(cx)
}
}
#[cfg(test)]
mod io_tests {
use super::*;
use async_std::io::ReadExt;
use async_std::prelude::FutureExt;
use async_std::{
io::{copy, Cursor},
net::TcpStream,
};
use std::time::Duration;
#[async_std::test]
async fn fail_read_after_timeout() -> io::Result<()> {
let mut output = b"______".to_vec();
let io = PendIo;
let mut io = PrudentIo::new(Duration::from_millis(5), io);
let mut io = Pin::new(&mut io);
insta::assert_debug_snapshot!(io.read(&mut output[..]).timeout(Duration::from_secs(1)).await,@"Ok(io::Err(timeou))");
Ok(())
}
#[async_std::test]
async fn timeout_expires() {
let later = delay(Duration::from_millis(1)).expect("some").await;
insta::assert_debug_snapshot!(later,@r"()");
}
/// Mock IO always pending
struct PendIo;
impl io::Read for PendIo {
fn poll_read(
self: Pin<&mut Self>,
_cx: &mut Context<'_>,
_buf: &mut [u8],
) -> Poll<futures_io::Result<usize>> {
Poll::Pending
}
}
impl io::Write for PendIo {
fn poll_write(
self: Pin<&mut Self>,
_cx: &mut Context<'_>,
_buf: &[u8],
) -> Poll<futures_io::Result<usize>> {
Poll::Pending
}
fn poll_flush(self: Pin<&mut Self>, _cx: &mut Context<'_>) -> Poll<futures_io::Result<()>> {
Poll::Pending
}
fn poll_close(self: Pin<&mut Self>, _cx: &mut Context<'_>) -> Poll<futures_io::Result<()>> {
Poll::Pending
}
}
}发布于 2022-02-08 19:32:06
异步超时工作如下:
similar.
poll进入超时,它检查超时是否过期。
Ready并完成。
cx.waker().wake()的正确时间过去时,或者在适当的时间内调用#4的回调,即调用适当的waker中的wake(),指示运行库再次调用poll。
poll将返回Ready。好了!代码的问题是从poll()实现:self.expired = delay(self.timeout);内部创建延迟。但是,在不轮询超时的情况下返回Pending甚至一次。这样,就不会在任何地方注册回调来调用Waker。没有懦夫,没有暂停。
我认为有几种解决办法:
A.不要将PrudentIo::expired初始化为None,而是直接在构造函数中创建timeout。这样,超时将始终在io之前被轮询至少一次,并且会被唤醒。但是,您将始终创建一个超时,即使它实际上并不需要。
B。创建timeout时,执行递归轮询:
Poll::Pending => {
if self.expired.is_none() {
// No data, start checking for a timeout
self.expired = delay(self.timeout);
return self.poll_read(cx, buf);
}这将调用io两次,这是不必要的,因此可能不是最优的。
C.在创建超时后添加对轮询的调用:
Poll::Pending => {
if self.expired.is_none() {
// No data, start checking for a timeout
self.expired = delay(self.timeout);
self.expired.as_mut().unwrap().as_mut().poll(cx);
}也许您应该匹配轮询的输出,以防它返回Ready,但是嘿,这是一个新的超时,它可能还在等待,而且它似乎工作得很好。
发布于 2022-03-04 08:19:11
// This is another solution. I think it is better.
impl<IO: io::AsyncRead + Unpin> io::AsyncRead for PrudentIo<IO> {
fn poll_read(
self: Pin<&mut Self>,
cx: &mut Context<'_>,
buf: &mut [u8],
) -> Poll<io::Result<usize>> {
let this = self.get_mut();
let io = Pin::new(&mut this.io);
if let Poll::Ready(res) = io.poll_read(cx, buf) {
return Poll::Ready(res);
}
loop {
if let Some(expired) = this.expired.as_mut() {
ready!(expired.poll(cx));
this.expired.take();
return Poll::Ready(Err(io::ErrorKind::TimedOut.into()));
}
let timeout = Timer::after(this.timeout);
this.expired = Some(timeout);
}
}
}发布于 2022-03-04 03:27:09
// 1. smol used, not async_std.
// 2. IO should be 'static.
// 3. when timeout, read_poll return Poll::Ready::Err(io::ErrorKind::Timeout)
use {
smol::{future::FutureExt, io, ready, Timer},
std::{
future::Future,
pin::Pin,
task::{Context, Poll},
time::Duration,
},
};
// --
pub struct PrudentIo<IO> {
expired: Option<Pin<Box<dyn Future<Output = io::Result<usize>>>>>,
timeout: Duration,
io: IO,
}
impl<IO> PrudentIo<IO> {
pub fn new(timeout: Duration, io: IO) -> Self {
PrudentIo {
expired: None,
timeout,
io,
}
}
}
impl<IO: io::AsyncRead + Unpin + 'static> io::AsyncRead for PrudentIo<IO> {
fn poll_read(
self: Pin<&mut Self>,
cx: &mut Context<'_>,
buf: &mut [u8],
) -> Poll<io::Result<usize>> {
let this = self.get_mut();
loop {
if let Some(expired) = this.expired.as_mut() {
let res = ready!(expired.poll(cx))?;
this.expired.take();
return Ok(res).into();
}
let timeout = this.timeout.clone();
let (io, read_buf) = unsafe {
// Safety: ONLY used in poll_read method.
(&mut *(&mut this.io as *mut IO), &mut *(buf as *mut [u8]))
};
let fut = async move {
let timeout_fut = async {
Timer::after(timeout).await;
io::Result::<usize>::Err(io::ErrorKind::TimedOut.into())
};
let read_fut = io::AsyncReadExt::read(io, read_buf);
let res = read_fut.or(timeout_fut).await;
res
}
.boxed_local();
this.expired = Some(fut);
}
}
}
impl<IO: io::AsyncWrite + Unpin> io::AsyncWrite for PrudentIo<IO> {
fn poll_write(
mut self: Pin<&mut Self>,
cx: &mut Context<'_>,
buf: &[u8],
) -> Poll<io::Result<usize>> {
Pin::new(&mut self.io).poll_write(cx, buf)
}
fn poll_flush(mut self: Pin<&mut Self>, cx: &mut Context<'_>) -> Poll<io::Result<()>> {
Pin::new(&mut self.io).poll_flush(cx)
}
fn poll_close(mut self: Pin<&mut Self>, cx: &mut Context<'_>) -> Poll<io::Result<()>> {
Pin::new(&mut self.io).poll_close(cx)
}
}https://stackoverflow.com/questions/71039085
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